Since \(f\) is injective, \(f'\) is nonvanishing in \(\Omega \coloneqq{\left\lvert {z} \right\rvert} \leq r\). A computation: \begin{align*} \mu(f({\mathbb{D}}_r)) &= \int_{{\mathbb{D}}_r} {\left\lvert {f'(z)} \right\rvert}\,dz\\ &= \int_{{\mathbb{D}}_r} f'(z) \mkern 1.5mu\overline{\mkern-1.5muf'(z)\mkern-1.5mu}\mkern 1.5mu \,dz\\ &= \int_{{\mathbb{D}}_r} \qty{\sum_{j\geq 1} jc_j z^{j-1} } \mkern 1.5mu\overline{\mkern-1.5mu \qty{\sum_{k\geq 1} kc_k z^{k-1}}\mkern-1.5mu}\mkern 1.5mu \,dz\\ &= \int_{{\mathbb{D}}_r} \qty{\sum_{j\geq 1} j c_j z^{j-1} } { \qty{\sum_{k\geq 1} k \mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu^{k-1} }} \,dz\\ &= \int_{{\mathbb{D}}_r} \sum_{j, k\geq 1} jk c_j \mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu z^{j-1}\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu^{k-1}\,dz\\ &= \int_0^R\int_0^{2\pi} \sum_{j, k\geq 1} jk c_j \mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu (re^{it})^{j-1} (re^{-it})^{k-1} r\,dr\,dt\\ &= \int_0^R\int_0^{2\pi} \sum_{j, k\geq 1} jk c_j \mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu r^{j+k-1} e^{i(j-k)t} \,dr\,dt\\ &= \int_0^R \sum_{k\geq 1} k^2 {\left\lvert {c_k} \right\rvert}^2 r^{2k-1} \cdot 2\pi \,dt\\ &= \sum_{k\geq 1} k^2 {\left\lvert {c_k} \right\rvert}^2 {r^{2k} \over 2k}\Big|_{r=0}^R \\ &= \pi \sum_{k\geq 1} k {\left\lvert {c_k} \right\rvert}^2 R^{2k} .\end{align*}

See above solution: all goes identically up until the integral over \(r\) values, just replace \(\int_0^R\) with \(\int_r^R\).

With some substitution, one can compute \begin{align*} {\left\lvert {x_n - x_{n-1}} \right\rvert} = {\left\lvert {{1\over 2} x_{n-1} + {1\over 2} x_{n-2} - x_{n-1}} \right\rvert} = {1\over 2} {\left\lvert {x_{n-1} - x_{n-2}} \right\rvert} ,\end{align*} which holds for all \(n\). This is enough to show that the sequence is contractive, i.e.  \begin{align*} {\left\lvert {x_n - x_{n-1}} \right\rvert} = c {\left\lvert {x_{n-1} - x_{n-2}} \right\rvert} && c\in (0, 1) .\end{align*}

Apply this recursively yields \begin{align*} {\left\lvert {x_n - x_{n-1}} \right\rvert} = \qty{1\over 2}^{n-1} {\left\lvert {b-a} \right\rvert} \overset{n\to\infty}\longrightarrow 0 ,\end{align*} since \({\left\lvert {b-a} \right\rvert}\) is a constant. So in fact \(x_n\) is convergent and thus Cauchy convergent.

Note: to compare \({\left\lvert {x_i - x_j} \right\rvert}\) directly, assume \(i>j\) and apply the above estimate \(i-j+1\) on \({\left\lvert {x_i - x_{i-1}} \right\rvert}, {\left\lvert {x_{i-1} - x_{i-2}} \right\rvert}, \cdots\) to reduce to this case. This yields something like \begin{align*} {\left\lvert {x_i - x_j} \right\rvert} = \qty{1\over 2}^{i-j+1}{\left\lvert {x_{j} - x_{j-1}} \right\rvert} = \qty{1\over 2}^{i-j+1} \qty{1\over 2}^{j-1} {\left\lvert {b-a} \right\rvert}\to 0 .\end{align*} One could equivalently use the triangle inequality and a partial geometric sum to write \begin{align*} {\left\lvert {x_i - x_j} \right\rvert} \leq \sum_{j\leq k \leq i-1} {\left\lvert {x_{k+1} - x_{k}} \right\rvert} \implies {\left\lvert {x_i - x_j} \right\rvert} \leq c^j\qty{1\over 1-c}{\left\lvert {b-a} \right\rvert} .\end{align*}

Computing its limit: the usual trick of setting \(L \coloneqq\lim x_n = \lim x_{n-1} = \lim x_{n-2}\) only yields \(L = {L + L \over 2}\) here, and thus no information. Instead assume \(x_n = r^n\) is geometric, then \begin{align*} 2x_n - x_{n-1} - x_{n-2} = 0 \implies 2r^n - r^{n-1} - r^{n-2} = 0 \implies 2r^2 - r - 1 = 0 \iff (2r+1)(r-1) = 0 \implies r = -1/2, 1 .\end{align*} Write a general solution as \begin{align*} x_n = c_1 (-1/2)^n + c_2 (1)^n = c_1 (-1/2)^n + c_2 ,\end{align*} and solve for initial conditions: \begin{align*} x_0: \quad a &= c_1 + c_2 \\ x_1: \quad b &= (-1/2)c_1 + c_2 \\ \\ \implies { \begin{bmatrix} {1} & {1} \\ {-1/2} & {1} \end{bmatrix} } \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} &= \begin{bmatrix} a \\ b \end{bmatrix} \\ \implies \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} &= {1\over 1 + (1/2)} { \begin{bmatrix} {1} & {-1} \\ {1/2} & {1} \end{bmatrix} } \begin{bmatrix} a \\ b \end{bmatrix} \\ &= \qty{1\over 3} { \begin{bmatrix} {2} & {-2} \\ {1} & {2} \end{bmatrix} } \begin{bmatrix} a \\ b \end{bmatrix} \\ &= \qty{1\over 3} \begin{bmatrix} 2a-2b \\ a+b \end{bmatrix} .\end{align*}

So the general solution is \begin{align*} x_n = {2\over 3}(a-b) \qty{-1\over 2}^n + {1\over 3}(a+b)\overset{n\to \infty}\longrightarrow \qty{1\over 3}(a+b) .\end{align*}

Fix \({\varepsilon}>0\), we need to find a \(\delta = \delta({\varepsilon})\) such that \begin{align*} {\left\lvert {x-y} \right\rvert}<\delta \implies {\left\lvert {f(x) - f(y)} \right\rvert} < {\varepsilon}&& \forall x, y\in {\mathbb{R}} .\end{align*} Use that \(\lim_x\to \pm \infty f(x) = 0\) to choose \(M\gg 0\) such that \begin{align*} {\left\lvert {x} \right\rvert} \geq M \implies {\left\lvert {f(x)} \right\rvert} \leq {\varepsilon}/2 ,\end{align*} then \begin{align*} {\left\lvert {x} \right\rvert}, {\left\lvert {y} \right\rvert} \geq M \implies {\left\lvert {f(x) - f(y)} \right\rvert} \leq {\left\lvert {f(x)} \right\rvert} + {\left\lvert {f(y)} \right\rvert} \leq {\varepsilon} .\end{align*} So in this region choose (say) \(\delta_1 < {\varepsilon}\) to ensure that \(B_\delta(x), B_\delta(y) \subseteq [-M, M]^c\). On \([-M, M]\), note that this region is compact and \(f\) continuous on a compact set implies uniformly continuous. So use this to choose \(\delta_2 = \delta_2({\varepsilon})\) in this region to ensure \({\left\lvert {f(x) - f(y)} \right\rvert} < {\varepsilon}\).

This handles the cases \(x, y \in (M, M)^c\), or \(x,y\in [M, M]\), so it only remains to handle \(x\in [M, M]\) and \(y\in (M, M)^c\) (wlog, relabeling \(x,y\) if necessary). In this case, use the triangle inequality: \begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} &= {\left\lvert {f(x) - f(M) + f(M) -f(y)} \right\rvert} \\ &\leq {\left\lvert {f(x) - f(M)} \right\rvert} + {\left\lvert {f(M) -f(y)} \right\rvert} \\ &\leq {\varepsilon}+ {\left\lvert {f(M)} \right\rvert} + {\left\lvert {f(y)} \right\rvert} \\ &\leq {\varepsilon}+ {\varepsilon}+ {\varepsilon} ,\end{align*} where we’ve used that \(M, y\in (M, M)^c\) to apply the first bound and \(M, x\in [M, M]\) to apply the second.

A direct computation: fix \({\varepsilon}>0\) and suppose \({\left\lvert {z-w} \right\rvert} < R\). Then \begin{align*} {\left\lvert {z^2-w^2} \right\rvert} &= {\left\lvert {z-w} \right\rvert}{\left\lvert {z+w} \right\rvert} \\ &\leq \delta \qty{{\left\lvert {z} \right\rvert} + {\left\lvert {w} \right\rvert}} \\ &\leq \delta \cdot 2R ,\end{align*} so choose \(\delta < { {\varepsilon}\over 2R}\) to get uniform continuity on \({\mathbb{D}}_{R/2}(0)\).

To see \(f\) can’t be uniformly continuous on \({\mathbb{C}}\), take \({\varepsilon}\coloneqq c\) any constant and suppose the appropriate \(\delta\) exists. We’ll look for a bad pair of \(z, w\), so take \(w = z + {1\over 2}\delta\). This would imply \begin{align*} {\left\lvert {z^2 - w^2} \right\rvert} &= {\left\lvert {z^2 - (z+\delta)^2} \right\rvert} \\ &= {\left\lvert {-2z\delta - \delta^2} \right\rvert} \\ &= {\left\lvert {2z\delta + \delta^2} \right\rvert} \\ &= \delta {\left\lvert {2z + \delta} \right\rvert} \\ &\overset{{\left\lvert {z} \right\rvert}\to\infty}\longrightarrow\infty ,\end{align*} using the \(\delta = \delta({\varepsilon})\) can’t depend on \(z\) or \(w\), and is thus constant in this expression. This contradicts that \({\left\lvert {z^2-w^2} \right\rvert} < {\varepsilon}= c < \infty\).

The standard example: \begin{align*} f(x) \coloneqq \begin{cases} x^2\sin\qty{1\over x} & x\neq 0 \\ 0 & x=0. \end{cases} .\end{align*}

Away from zero, this is clearly differentiable since we can just compute the derivative by the chain rule. It turns out that \begin{align*} f'(x) = \begin{cases} 2x\sin\qty{1\over x} + x^2 \cos\qty{1\over x}\qty{-1\over x^2} = 2x\sin\qty{1\over x} - \cos\qty{1\over x} & x\neq 0 \\ 0 & x=0. \end{cases} .\end{align*} Here we check differentiability and compute the derivative at \(x=0\) directly: \begin{align*} {f(x) - f(0) \over x-0} = {x^2\sin\qty{1\over x} - 0 \over x-0} = x\sin\qty{1\over x} \overset{x\to 0}\longrightarrow 0 ,\end{align*} using that \(-x \leq {\left\lvert {x\sin \qty{1\over x}} \right\rvert}\leq x\).

But now notice that the \(\cos\qty{1\over x}\) term in \(f'\) isn’t enveloped by an \(x^c\) term, so \(\lim_{x\to 0} f'(x)\) does not exist for oscillatory reasons:

In particular, \(\lim_{x\to 0}f'(x) \neq f'(0) = 0\).

Uniformly convergent means that \({\left\lVert {g_i - g_j} \right\rVert}_{\infty} \to 0\), so \(\sup_{x\in X}{\left\lvert {g_i(x)-g_j(x)} \right\rvert} \overset{i, j\to\infty}\longrightarrow 0\). We want to show that given \({\varepsilon}\) we can find \(N_0\) such that \(i, j > N_0\) yields \begin{align*} \sup_{x\in X}{\left\lvert { f\circ g_i(x) - f\circ g_j(x) } \right\rvert} < {\varepsilon} .\end{align*}

Fix \({\varepsilon}> 0\), then choose \(\delta_1 = \delta_1({\varepsilon})\) by uniform continuity of \(f\) to guarantee \begin{align*} {\left\lvert {y_1 - y_2} \right\rvert} \leq \delta_1 \implies {\left\lvert {f(y_1) - f(y_2) } \right\rvert} < {\varepsilon}\, \forall y_1, y_2\in X .\end{align*} Now by uniform convergence of \(\left\{{g_n}\right\}\), choose \(N_0 = N_0(\delta_1)\) such that \begin{align*} i, j \geq N_0 \implies {\left\lvert { g_i(x) - g_j(x) } \right\rvert} < \delta_1 \, \forall x\in X .\end{align*}

Now writing \(y_1 \coloneqq g_i(x), y_2 \coloneqq g_j(x)\), choose \(i, j > N_0\) yields \begin{align*} {\left\lvert {y_1 - y_2} \right\rvert} \coloneqq{\left\lvert {g_i(x) - g_j(x) } \right\rvert} < \delta_1 \\ \implies {\left\lvert {f(y_1) - f(y_2)} \right\rvert} \coloneqq{\left\lvert {f(g_i(x)) - f(g_j(x))} \right\rvert} < {\varepsilon} ,\end{align*} and taking the supremum over \(x\in X\) preserves the inequality since \(\delta_1\) and consequently \(N_0\) only depend on \({\varepsilon}\).

\(\implies\): Fix \({\varepsilon}>0\) and choose \(\delta = \delta({\varepsilon})\) to get a bound corresponding to \({\varepsilon}/2\), then for all \(x,y\) with \({\left\lvert {x-y} \right\rvert} < \delta\) on \([a, b]\), we have \begin{align*} {\left\lvert {f'(x) - f'(y) } \right\rvert} \leq {\left\lvert {f'(x) - {f(x) - f(y) \over x- y} } \right\rvert} + {\left\lvert { {f(x) - f(y) \over x-y} - f'(y)} \right\rvert} < {\varepsilon} .\end{align*} This uses uniformity to bound by \({\varepsilon}/2\) the terms involving \(f'(x)\) and \(f'(y)\) respectively. So \(f'\) is in fact uniformly continuous on \([a, b]\).

\(\impliedby\): Let \({\varepsilon}> 0\) and \(x,y\in [a, b]\) be arbitrary. Then by the MVT, we can a \(\xi\in [x, y]\) with \(f'(\xi)(x-y) = f(x) - f(y)\). Then use continuity of \(f'\) to choose \(\delta = \delta({\varepsilon}, x, y)\) so that \({\left\lvert {x-y} \right\rvert} < \delta \implies {\left\lvert {f(x) - f(y)} \right\rvert} < {\varepsilon}\), and note that \({\left\lvert {x-\xi} \right\rvert} \leq {\left\lvert {x-y} \right\rvert} < \delta\), so \begin{align*} {\left\lvert { {f(x) - f(y) \over x-y } - f'(y) } \right\rvert} = {\left\lvert { f'(\xi) - f'(y)} \right\rvert} < {\varepsilon} .\end{align*}

Define a function \begin{align*} d: A \times B &\to {\mathbb{R}}\\ (x, y) &\mapsto {\left\lVert {x- y} \right\rVert} .\end{align*} Then \(d\) is a continuous function on a compact topological space (where the product is compact by Tychonoff), and the extreme value theorem applies: \(d\) attains its min/max for some pair \((a, b)\) in its domain.

Note that disjointness just guarantees that \({\left\lVert {a-b} \right\rVert}>0\), since \({\left\lVert {a-b} \right\rVert} = 0 \implies a=b\) and \(A \cap B = \emptyset\).

Use that \(X\) is connected iff \(\mathop{\mathrm{Hom}}_{{\mathsf{Top}}}(X, S^0) = \left\{{c_{-1}, c_1}\right\}\), i.e. every continuous map from \(X\to \left\{{-1, 1}\right\}\) is a constant map \(x \xrightarrow{c_{-1}} -1\) or \(x \xrightarrow{c_1} 1\). Let \(f: A\cup B \to S^0\) be arbitrary, and let \(f_1 \coloneqq{ \left.{{f}} \right|_{{A}} }\) and \(f_2 \coloneqq{ \left.{{f}} \right|_{{B}} }\). By connectedness of \(A\), \(f_1\) is a constant map, as is \(f_2\). On the intersection, for \(x\in A \cap B \neq \emptyset\), we have \(f_1(x) = f_2(x)\) since \(x\in A\) and \(x\in B\). So \(f_1\) and \(f_2\) are constant functions that must map to the same constant, so \(f\) is constant and this \(A\cup B\) is connected.

Let \({\varepsilon}>0\), we want to show that there exists an \(N_0\) such that \(n\geq N_0\) implies \({\left\lVert {f_n} \right\rVert}_\infty<{\varepsilon}\). Fix \(x\), by pointwise convergence pick \(M_x = M_x(x, {\varepsilon})\) so that \(n\geq M \implies {\left\lvert {f_n(x)} \right\rvert} < {\varepsilon}\). By continuity, this bound holds in some neighborhood \(U_x \ni x\). Produce a cover \(\left\{{U_x}\right\}_{x\in [0, 1]}\rightrightarrows[0, 1]\); by compactness produce a finite subcover \(\left\{{U_1, \cdots, U_m}\right\} \rightrightarrows[0, 1]\). Each \(U_i\) corresponds to some \(x_i\) and some \(M_{x_i}\), so choose \(N_0 > \max_{i\leq m} \left\{{M_{x_i}}\right\}\). Then \(n\geq N_0 \implies N\geq M_{x_i}\) for each \(i\), so \({\left\lvert {f_n(x)} \right\rvert} < {\varepsilon}\) for each \(x\in [0, 1]\) since \(x\in U_i\) for some \(i\). So \(\sup_{x\in X} {\left\lvert {f_n(x)} \right\rvert} = {\left\lVert {f_n} \right\rVert}_{\infty } < {\varepsilon}\).

See 3.2.12 of Understanding analysis 2ed. of Abbott. Show something stronger, that the following set is nonempty and open: \begin{align*} S \coloneqq\left\{{t\in {\mathbb{R}}{~\mathrel{\Big\vert}~}E \cap(-\infty, t), E \cap(t, \infty) \text{ are uncountable}}\right\} \subseteq {\mathbb{R}} .\end{align*} Write \begin{align*} S_- &\coloneqq\left\{{ t\in {\mathbb{R}}{~\mathrel{\Big\vert}~}E \cap(- \infty, t) \text{ is countable}}\right\} \\ S_+ &\coloneqq\left\{{ s\in {\mathbb{R}}{~\mathrel{\Big\vert}~}E \cap(s, \infty) \text{ is countable}}\right\} .\end{align*}

Note that \(S_- \neq {\mathbb{R}}\) since then we could write \(E = \displaystyle\bigcup_{n\in {\mathbb{Z}}} E \cap(- \infty, n)\) as a countable union of countable sets.

Claim: \(S = (\sup S_-,, \inf S_+)\).

???

Write \(U(f), L(f)\) for the upper and lower sums of \(f\), so for \(\Pi\) the collection of all partitions of \([0, 1]\), \begin{align*} U(f) \coloneqq\inf_{P\in \Pi} U(f, P) && U(f, P) \coloneqq\sum_{k=1}^n \sup_{x\in I_k}f(x) \cdot \mu(I_k) \\ L(f) \coloneqq\sup_{P\in \Pi} L(f, P) && L(f, P) \coloneqq\sum_{k=1}^n \inf_{x\in I_k} f(x) \cdot \mu(I_k) .\end{align*}

Note that integrability of \(f\) is equivalent to \begin{align*} \forall {\varepsilon}\exists P \text{ such that } U(f, P) - L(f, P) < {\varepsilon}\\ \iff \sum_{k=1}^n \qty{ \sup_{x\in I_k} f(x) - \inf_{x\in I_k} f(x)} \mu(I_k) < {\varepsilon} .\end{align*}

\begin{align*} {\left\lvert {x^n - y^n} \right\rvert} = {\left\lvert {y-x} \right\rvert}{\left\lvert {\sum_{1\leq k \leq n} x^k y^{n-k}} \right\rvert} \leq n M^{n-1}{\left\lvert {y-x} \right\rvert} \overset{y\to x}\longrightarrow 0 .\end{align*}

\begin{align*} x^{1\over n} - y^{1\over n} \leq (x-y)^{1\over n} \overset{x\to y}\longrightarrow 0 ,\end{align*} using \((a+b)^m \geq a^m + b^m\)

Apply the MVT: \begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} = {\left\lvert {f(\xi)} \right\rvert} {\left\lvert {x-y} \right\rvert} \overset{y\to x}\longrightarrow 0 .\end{align*}

Check \(\sup f = 1\) and \(\inf f = 0\) on every sub-interval, so \(L(f, P) = 0\) and \(U(f, P) = 1\) for every partition \(P\) of \([0, 1]\).

Discontinuity: #todo

Note that \(g\) is entire by Morera’s theorem, since \(0 = \int_T f_n \to \int_T g\) by uniform convergence and the \(f_n\) are holomorphic. By Cauchy’s theorem, up to a constant we have \begin{align*} f_n(z) = \oint_T {f_n(\xi) \over \xi - z}\,d\xi g(z) = \oint_T {g(\xi) \over \xi - z}\,d\xi ,\end{align*} Thus fixing \(K\) and \({\varepsilon}\), for any \(T \subseteq K\) containing \(z\), ` \begin{align*} {\left\lvert {f_n(z) - g(z)} \right\rvert} &= {\left\lvert { \oint_T {f_n(\xi) \over \xi - z},d\xi

• \oint_T {g(\xi) \over \xi - z},d\xi } \right\rvert} \ &\leq \oint_T {{\left\lvert {f_n(\xi) - g(\xi)} \right\rvert} \over \xi - z },d\xi\ &\leq \oint_T { \sup_{\xi\in T}{\left\lvert {f_n(\xi) - g(\xi)} \right\rvert} \over \xi - z },d\xi\ &\leq \oint_T { {\varepsilon}\over \xi - z },d\xi\ &= {\varepsilon}C \to 0 ,\end{align*} `{=html} where \(n = n({\varepsilon}, T)\) can be chosen to produce this \({\varepsilon}\) using that \(f_n\to g\) uniformly on \(T\). Taking a sup over the \(z\) enclosed by \(T\) on the LHS yields a bound on the open region enclosed by \(T\). Taking a union of all such \(T\) in \(K\) yields an open cover of \(K\), which by compactness has a finite subcover. This yields a finite collection \(\left\{{n = n({\varepsilon}, T_k)}\right\}_{k\leq N}\), and taking the maximum such \(n\) yields a uniform bound for all of \(K\).

Write \(g(z) \coloneqq\lim_{n\to\infty}f_n(z)\) for the pointwise limit, then \(g(z) = f(z)\) on a set with a limit point. By the identity principle, \(g\equiv f\) on \(\Omega\), making \(f\) the pointwise limit of the \(f_n\).

By Montel, locally uniformly bounded implies normal and locally equicontinuous. So \(\left\{{f_n}\right\}\) is normal, and thus has a locally uniformly convergent subsequence \(\left\{{f_{n_k}}\right\}\). Since singletons \(\left\{{z}\right\}\) are compact, \(f_{n_k}(z) \to g(z)\) pointwise, and by uniqueness of limits, \(\lim_{k\to\infty } f_{n_k} = g = f\) on any compact \(K \subseteq \Omega\).

It remains to show that the original sequence \(\left\{{f_n}\right\}\) converges locally uniformly to \(f\), not just the subsequence. Suppose not, then there exists a compact \(K \subseteq \Omega\) and \({\varepsilon}>0\) so that \({\left\lVert {f_n - f} \right\rVert}_{K, \infty} > {\varepsilon}\) for infinitely many \(n\). This produces a subsequence \(\left\{{f_{n_j}}\right\}\) with \({\left\lVert {f_{n_j} - f} \right\rVert} > {\varepsilon}\) for all \(j\). However, since \({\mathcal{F}}\) was normal, every subsequence has a locally uniformly convergent subsequence, so this has a further subsequence \(f_{n_{j'}}\) uniformly converging to \(f\), a contradiction.

Let \(K\) be compact, where \(z\in K\implies {\left\lvert {z} \right\rvert} \leq R\) for some constant \(R\). For the remainder of the problem, we only work in \(K\).

Why these claims imply the result:

If \(f_n(z)\to z\) uniformly, both are uniformly bounded, and \(e^z\) is uniformly continuous, then \(e^{f(z)}\to e^z\) uniformly.

\(\implies\):

• Fix \(r \in (0, 1)\) and let \(\gamma = \left\{{{\left\lvert {z} \right\rvert} = r}\right\}\). This is compact, so \(f_n\to f\) uniformly on \(\gamma\): \begin{align*} \int_\gamma {\left\lvert {f_n(z) - f(z) } \right\rvert} \,dz &\leq\int_\gamma \sup_{w\in \gamma } {\left\lvert {f_n(w) - f(w) } \right\rvert} \,dz\\ &\leq\int_\gamma {\left\lVert {f_n(w) - f(w) } \right\rVert}_{\infty} \,dz\\ &= {\left\lVert {f_n(w) - f(w) } \right\rVert}_{\infty} \int_\gamma \,dz\\ &= {\left\lVert {f_n(w) - f(w) } \right\rVert}_{\infty} \mathop{\mathrm{length}}(\gamma) \\ &\overset{n\to\infty}\longrightarrow 0 .\end{align*}

\(\impliedby\):

• Let \(K\) be compact, then choose \(\gamma\) enclosing but not intersecting \(K\).

• Since \(\gamma, K\) are disjoint compact sets, define \(M \coloneqq\inf \left\{{{\left\lvert {z-\xi} \right\rvert} {~\mathrel{\Big\vert}~}z\in K, \xi\in \gamma}\right\}\), the \(0<M<\infty\).

• Apply Cauchy’s formula to the function \(F_n(z) \coloneqq f_n(z) - f(z)\), where we want to show \({\left\lvert {F_n(z)} \right\rvert} < {\varepsilon}\): \begin{align*} F_n(z) &= {1\over 2\pi i} \int_\gamma { F_n(\xi) \over z-\xi} \,d\xi\\ \implies {\left\lvert {f_n(z) - f(z) } \right\rvert} &\leq {1\over 2\pi }\int_\gamma {\left\lvert {f_n(\xi) - f(\xi) \over z-\xi} \right\rvert} \,d\xi\\ &\leq {1\over 2\pi} \int_\gamma {{\left\lvert { f_n(\xi) - f(\xi) } \right\rvert} \over M} \,d\xi\\ &\leq {1\over 2\pi M} \int_\gamma {{\left\lvert { f_n(\xi) - f(\xi) } \right\rvert} } {\left\lvert {\,d\xi} \right\rvert} \\ ,\end{align*} where by hypothesis we can bound this integral by an \({\varepsilon}\). So given \({\varepsilon}\), choose \(n\) large enough to bound the integral as above by some \({\varepsilon}\) depending only on \(n\) and not on \(z\). Taking \(\sup\) of both sides yields \({\left\lVert {f_n - f} \right\rVert}_{\infty, K} \leq {{\varepsilon}\over 2\pi M}\), so \(f_n\to f\) uniformly on \(K\).

No: polynomials are holomorphic and the uniform limit of holomorphic functions is holomorphic. However, \(f(z) \coloneqq\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\) is continuous on \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu\) but not holomorphic, so can not be uniformly approximated by any sequence of polynomials.

The key insight: \begin{align*} \int_\gamma A \,dz &= \int_0^b A \cdot i \,dt&& z=x+it,\, \,dz= i\,dt\\ &=iA \int_0^b \,dt\\ &= iAb .\end{align*}

So now estimate the difference: \begin{align*} {\left\lvert { \int_{\gamma} f(z) \,dz- iAb } \right\rvert} &= {\left\lvert { \int_\gamma f(z) \,dz- \int_\gamma A \,dz} \right\rvert} \\ &= {\left\lvert { \int_\gamma \qty{ f(z) - A } \,dz} \right\rvert} \\ &\leq\int_\gamma {\left\lvert { f(z) - A } \right\rvert} \,dz\\ &\leq \sup_{z = x+iy\in \gamma} {\left\lvert {f(x+iy) - A} \right\rvert} \cdot \mathop{\mathrm{length}}(\gamma_x) \\ &\overset{x\to \infty}\longrightarrow 0 ,\end{align*} using that \(\mathop{\mathrm{length}}(\gamma_x) = b\) is constant.

Key observation: \begin{align*} iA\alpha = \int_\gamma {A\over z}\,dz .\end{align*} Why this is true: \begin{align*} \int_\gamma {A\over z}\,dz= \int_0^\alpha {1\over Re^{it}} iRe^{it}dt = \int_0^\alpha iA \,dt= iA\alpha .\end{align*}

Now estimate the difference:

\begin{align*} {\left\lvert { \int_\gamma f(z) \,dz- iA\alpha } \right\rvert} &= {\left\lvert { \int_\gamma f(z) \,dz- \int_\gamma {A\over z} \,dz} \right\rvert}\\ &= {\left\lvert {\int_\gamma f(z) - {A\over z} \,dz} \right\rvert} \\ &= {\left\lvert {\int_\gamma{zf(z) - A \over z} \,dz} \right\rvert} \\ &\leq \int_\gamma {\left\lvert {zf(z) - A \over z} \right\rvert} \,dz\\ &= \int_\gamma { {\left\lvert {zf(z) - A} \right\rvert} \over R} \,dz\\ &\leq {1\over R } \int_\gamma {\left\lVert {zf(z) - A} \right\rVert}_{\infty, \gamma} \,dz\\ &= {{\varepsilon}\over R}\cdot \mathop{\mathrm{length}}(\gamma) \\ &= {{\varepsilon}\over R} \cdot R\alpha \\ &= {\varepsilon}\alpha \\ &\overset{R\to\infty}\longrightarrow 0 .\end{align*}

General strategy: each has two expansions, so just compute them all and pick appropriate ones for regions afterwards.

For \(1\over z-3\): \begin{align*} {1\over z-3} &= -{1\over 3}{1\over 1- {z\over 3}} = -{1\over 3}\sum_{k\geq 0}3^{-k}z^k && {\left\lvert {z} \right\rvert} < 3 \\ &= {1\over z} {1\over 1 - {3\over z}} = z^{-1}\sum_{k\geq 0} 3^k z^{-k} && {\left\lvert {z} \right\rvert} > 3 .\end{align*}

For \(1\over 1-z^2\): \begin{align*} {1\over 1-z^2} &= \sum_{k\geq 0} z^{2k} && {\left\lvert {z} \right\rvert} < 1 \\ &= {1\over z^2} {-1\over 1- z^{-2}} = -z^{-2}\sum_{k\geq 0}z^{-2k} && {\left\lvert {z} \right\rvert} > 1 .\end{align*}

So take \begin{align*} 0 < {\left\lvert {z} \right\rvert} < 1 && f(z) &= \sum_{k\geq 0}z^{2k} - {1\over 3}\sum_{k\geq 0} 3^{-k}z^k \\ 1 < {\left\lvert {z} \right\rvert} < 3 && f(z) &= -z^{-2} \sum_{k\geq 0}z^{-2k} - {1\over 3}\sum_{k\geq 0} 3^{-k}z^k \\ 3 < {\left\lvert {z} \right\rvert} < \infty && f(z) &= -z^{-2} \sum_{k\geq 0}z^{-2k} + z^{-1}\sum_{k\geq 0}3^k z^{-k} .\end{align*}

Idea: use Cauchy’s integral formula to get a series in \((z-z_0)\). \begin{align*} f(z) &= \int {f(\xi) \over \xi -z} \,d\xi\\ &= \int f(\xi) \qty{ 1\over \xi - (z - z_0) - z_0 } \,d\xi\\ &= \int { f(\xi) \over\xi - z_0} \qty{ 1\over 1-w } \,d\xi&& w\coloneqq{z-z_0 \over \xi - z_0} \\ &= \int { f(\xi) \over\xi - z_0} \sum_{k\geq 0} w^k \,d\xi\\ &= \sum_{k\geq 0} \qty{\int {f(\xi) \over \xi - z_0} \,d\xi} w^k\\ &= \sum_{k\geq 0} \qty{\int {f(\xi) \over \xi - z_0} \,d\xi} w^k\\ &= \sum_{k\geq 0} \qty{\int {f(\xi) \over (\xi - z_0)^{k+1} } \,d\xi} (z-z_0)^k ,\end{align*} where we’ve integrated over a curve contained in \(D\) the disc of convergence, and that the power series for \(f\) converges uniformly on \(D\) to commute the sum and integral.

Part a: Take \(f(z) \coloneqq\displaystyle\sum n^{-2}z^n\), which converges absolutely for \({\left\lvert {z} \right\rvert}=1\) by the comparison test.

Part b: Take \(f(z) \coloneqq{1\over 1+z} = \sum_{k\geq 0} (-1)^k z^k\), then \(f(1) = 2\) by analytic continuation of the series at \(z=1\). Then \(a_k = (-1)^k\),

Part c: ??? Not clear if this is true, take \(f(z) = \sum z^n/n^2\).

Part 1: This series does not have small tails: writing \(c_n \coloneqq n z^n\) we have \({\left\lvert {c_n} \right\rvert} = {\left\lvert {nz^n} \right\rvert} = {\left\lvert {n} \right\rvert}\to \infty\) when \({\left\lvert {z} \right\rvert} = 1\).

Part 2: This converges absolutely and absolute convergence implies convergence: \begin{align*} {\left\lvert {\sum n^{-2} z^n} \right\rvert} \leq \sum {\left\lvert {n^{-2}z^n} \right\rvert} = \sum n^{-2} < \infty .\end{align*}

Part 3: Write \(f(z) = \sum_{k\geq 1} k^{-1}z^k\). The value \(f(1)\) is the harmonic series, which we know diverges from undergraduate Calculus. For \(z\neq 1\), apply summation by parts with \(a_k \coloneqq k^{-1}\) and \(b_k \coloneqq z^k\), so

• \(a_N = N^{-1}\)
• \(a_M = M^{-1}\)
• \(B_N = \sum_{k\leq N} z^k = {1-z^{N+1} \over 1-z}\)
• \(B_M = \sum_{k\leq M} z^k\)
• \(a_{n+1} - a_n = (n+1)^{-1}+ n^{-1}= - (n(n+1))^{-1}\)

Note that \({\left\lvert {B_N} \right\rvert} \leq C_z \coloneqq{2\over {\left\lvert {1-z} \right\rvert} }\) for any \(N\), since \({\left\lvert {z} \right\rvert} = 1\) is on \(S^1\) and the maximum distance between two points on \(S^1\) is 2. Moreover \(C_z < \infty\) when \(z\neq 1\).

Applying the formula: ` \begin{align*} {\left\lvert {\sum_^N n^{-1}z^n } \right\rvert} &\leq {\left\lvert { N^{-1}B_N

• M^{-1}B_{M-1}
• \sum_^{N-1} \left[ -(n(n+1))^{-1}B_n \right] } \right\rvert}\ &\leq N^{-1}C_z + M^{-1}C_z + \sum_{M\leq n \leq N-1} C_z \qty{1\over n^2 + n}\ &\leq C_z\qty{N^{-1}+ M^{-1}+ \sum_{M\leq n \leq N-1} n^{-2}} \ &\overset{M, N\to\infty}\longrightarrow 0 ,\end{align*} `{=html} where we’ve used the triangle inequality and convergence of \(\sum n^{-2}\). By the Cauchy criterion for sums, \(f(z)\) converges pointwise for \(z\neq 1\).

\(\impliedby\): Let \(z\in U\) be fixed. Since \(f\) is holomorphic at \(z\) and \(f'(z)\neq 0\), consider \(f(x, y)\) and its Jacobian as a real-valued function: \begin{align*} D_f = { \begin{bmatrix} {u_x} & {u_y} \\ {v_x} & {v_y} \end{bmatrix} } \implies \operatorname{det}(D_f) = u_x v_y - v_x u_y = u_x^2 + v_x^2 = {\left\lvert {f'} \right\rvert}^2 > 0 ,\end{align*} so the derivative matrix is invertible at \(z\). Applying the inverse function theorem yields that \(f\) is a smooth diffeomorphism on some neighborhood \(N\ni p\), and in particular is bijective on \(N\).

\(\not\impliedby\): If \(f'(z) = 0\) for some \(z\), then we claim that \(f\) can not be injective. Equivalently, injectivity of \(f\) implies \(f'\neq 0\). Suppose \(f\) is holomorphic at \(z_0\) but \(f'(z_0)=0\). Write \(h(z) \coloneqq f(z) - f(z_0)\), which has a zero \(z_0\) of some order \(k\geq 2\). For a disc \(D\) small enough about \(z_0\) avoiding the other (isolated) zeros of \(h\) and \(f'\), for any \(p\) in a neighborhood of \(z_0\) and contained in \(D\), \begin{align*} \int_{{{\partial}}D} {f'(\xi) \over f(\xi) - p} \,d\xi = {\sharp}Z(f(z) - p) ,\end{align*} using the argument principle and that \((f(\xi) - p)' = f'(\xi)\). But for \(D\) small enough, \({\sharp}Z(f(z) - p) = {\sharp}Z(f(z) - f(z_0)) = k\) by Rouché, so there are \(k\) solutions to \(f(z) = p\). Since \((f(z) - p)' \neq 0\) in \(D\), none of these can be repeated roots, so these \(k\) solutions are distinct, forcing \(f\) to be \(k\)-to-one and fail injectivity.

Expanding on the Rouché argument: set \(c \coloneqq\inf_{z\in D} {\left\lvert {f(z) - w_0} \right\rvert}\), then for \(D'\) of radius \(c\), set

• \(F(z) \coloneqq(f(z) - z_0) - (f(z) - p) = z-p\)
• \(G(z) = f(z) - z_0\)
• \((F+G)(z) = f(z) - p\)

Then \(F>G\) on \({{\partial}}D'\) will imply \(F, F+G\) have the same number of zeros within \(D'\), and this bound follows from \begin{align*} {\left\lvert {F(z)} \right\rvert} = {\left\lvert {z-p} \right\rvert} < c \leq {\left\lvert {f(z) - p } \right\rvert} ,\end{align*} where the first inequality is from making the disc small and the second from choosing \(c\) as an inf.

The easy check: \(f\) is differentiable iff \({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}_z f = 0\), but \begin{align*} { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}_z {\left\lvert {z} \right\rvert}^2 = { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}_z z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = z \neq 0 ,\end{align*} unless of course \(z=0\).

A more explicit check: check the limits. \begin{align*} {f(z) - f(0) \over z-0} = { {\left\lvert {z} \right\rvert}^2 \over z } = {z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \over z} = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \overset{z\to 0}\longrightarrow 0 ,\end{align*} so \(f\) is differentiable at \(w=0\). Now taking \(w = Re^{i\theta} \neq 0\), \begin{align*} {f(z) -f(w) \over z-w} = {{\left\lvert {z} \right\rvert}^2 - {\left\lvert {w} \right\rvert}^2 \over z - w} = {\qty{{\left\lvert {z} \right\rvert} + {\left\lvert {w} \right\rvert} } \qty{{\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert}} \over z-w } = {{\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert} \over z-w}\cdot \qty{{\left\lvert {z} \right\rvert} + {\left\lvert {w} \right\rvert}} .\end{align*} First let \(z\to w\) along \({{\partial}}{\mathbb{D}}_{R'}(0)\) where \(R' \coloneqq{\left\lvert {w} \right\rvert}\), so that the numerator vanishes and the limit is zero. Then let \(z\to w\) along the curve \(\left\{{tw{~\mathrel{\Big\vert}~}t\in [0, 1]}\right\}\), then \({\left\lvert {z} \right\rvert} = t {\left\lvert {w} \right\rvert}\), so the ratio becomes \begin{align*} {{\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert} \over z-w}\cdot \qty{{\left\lvert {z} \right\rvert} + {\left\lvert {w} \right\rvert}} &= {t{\left\lvert {w} \right\rvert} - {\left\lvert {w} \right\rvert} \over tw-w}\cdot \qty{t{\left\lvert {w} \right\rvert} + {\left\lvert {w} \right\rvert}} \\ &= {{\left\lvert {w} \right\rvert}\qty{t-1 } \over w(t-1)} \cdot {\left\lvert {w} \right\rvert}(t+1) \\ &= { {\left\lvert {w} \right\rvert}^2(t+1) \over w} \\ &= \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu(t+1) \\ &\overset{t\to 1}\to 2\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu ,\end{align*} which is nonzero is \(w\neq 0\).

Slick proof: use that no curve \(\gamma \subseteq {\mathbb{C}}\) is open in \({\mathbb{C}}\).

If \({\left\lvert {f} \right\rvert} = c = r^2\) for some \(r\), then the image of \(f\) is contained in the curve \({{\partial}}{\mathbb{D}}_r(0)\). Since \(f\) is holomorphic on the source domain \(\Omega\), \(f\) is an open map, so if \(f\) is nonconstant the \(f(\Omega)\) is open. But \(f(\Omega) \subseteq {{\partial}}{\mathbb{D}}_r(0)\) can not be open, so \(f\) must be constant.

The usual more direct proof: write \({\left\lvert {f(z)} \right\rvert} = u^2 = v^2 = r^2\). The claim is that both \(u\) and \(v\) are constant. Take partial derivatives and clear the factor of 2: \begin{align*} {\partial}_x: \quad uu_x + vv_x &= 0\\ {\partial}_y: \quad uu_y + vv_y &= 0 .\end{align*} Now apply CR: \(u_x= v_y, u_y=-v_x\), then \begin{align*} uu_x - vu_y &=0 \\ uu_y + vu_x &=0 .\end{align*} Multiply the first by \(u_x\) and the second by \(u_y\), then add \begin{align*} uu_x^2 - vu_y u_x &= 0 \\ uu_y^2 + vu_x u_y &=0 \\ \implies u(u_x^2 + u_y^2) &=0 .\end{align*} A similar calculation yields \(v(v_x^2 + v_y^2) = 0\), so If \(u(x,y) = v(x, y) = 0\) at any point, then \({\left\lvert {f} \right\rvert} = 0\) and \(f\equiv 0\), so we’re done. Otherwise, \(u,v\) do not simultaneously vanish, so we must have \begin{align*} 0 = u_x^2 + u_y^2 &\implies 0 = u_x = u_y \implies u \text{ constant }\\ 0 = v_x^2 + v_y^2 &\implies 0 = v_x = v_y \implies v \text{ constant } ,\end{align*} so \(f=u+iv\) is constant.

Write \(f=u+iv\), so \(u\equiv c\) is constant. Then \(u_x = u_y = 0\), and CR yields \(v_y = u_x = 0\) and \(v_y = -u_x = 0\), so \(v\) is constant, making \(f\) constant.

Slick proof: apply the open mapping theorem again, since \(\operatorname{Arg}(f) = \theta_0\) implies that \(\operatorname{im}(f) \subseteq \gamma\) for the curve \(\gamma \coloneqq\left\{{t e^{i\theta_0}{~\mathrel{\Big\vert}~}t\in {\mathbb{R}}}\right\}\) which has no open subsets.

Note that this implies that any \({\mathbb{R}}{\hbox{-}}\)valued holomorphic function is constant.

Write \(f=u+iv\) so \(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu = u +i\tilde v\) where \(\tilde v \coloneqq-v\). Then \(u, \tilde v\) are constant, so in particular \(\Re(f)\) is constant and by 2 \(f\) is constant.

It suffices to show that \(g(z) \coloneqq\mkern 1.5mu\overline{\mkern-1.5muf(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu\) satisfies CR. Write \(f=u+iv\), then \begin{align*} g(x, y) \coloneqq a(x, y) + ib(x, y) = u(x, -y) -i v(x, -y) ,\end{align*} so we want to show \(a_x = b_y\) and \(a_y = -b_x\). By the chain rule, \begin{align*} a_x &= {\partial}_x (x\mapsto u(x, -y)) = u_x \\ a_y &= {\partial}_x (y\mapsto u(x, y))\circ(y\mapsto -y) = -u_y \\ b_x &= {\partial}_x(x\mapsto -v(x, -y)) = -v_x \\ b_y &= {\partial}_x(y \mapsto - v(x, y))\circ(y\mapsto -y) = v_y .\end{align*} Now use CR for \(f\) to write \begin{align*} a_x &= u_x = v_y = b_y \\ a_y &= -u_y = v_x = -b_x .\end{align*}

Set \(g(z) \coloneqq(f(z^*))^* \coloneqq\mkern 1.5mu\overline{\mkern-1.5muf(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu\), we can then show \(g'\) exists: \begin{align*} \lim_{h\to 0} {g(z+h) - g(z) \over h} &\coloneqq\lim_{h\to 0} {f((z+h)^*)^* - f(z^*)^* \over h^{**}} \\ &= \lim_{h\to 0} {\qty{ f(z^* + h^*) - f(z^*) }^* \over h^{**}} \\ &= \lim_{h\to 0} \qty{ f(z^* + h^* ) - f(z^*) \over h^* }^* \\ &\coloneqq\qty{f'(z^*)}^* ,\end{align*} where we’ve used that \(w\mapsto w^*\) is continuous to commute a limit. So this limit exists, \(g\) is differentiable with \(g'(z) \coloneqq\mkern 1.5mu\overline{\mkern-1.5muf'(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu\).

Since \(f\) is analytic, take a Laurent expansion \(f(z) = \sum_{k\geq 0} c_k z^k\). Then \begin{align*} g(z) \coloneqq(f(z^*))^* = \qty{\sum_{k\geq 0} c_k \mkern 1.5mu\overline{\mkern-1.5muz^k\mkern-1.5mu}\mkern 1.5mu }^* = \sum_{k\geq 0} \mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu z^k ,\end{align*} making \(g\) analytic.

Part 1:

Write \begin{align*} x &= r\cos \theta \implies \operatorname{grad}_{r, \theta} x = {\left[ {\cos \theta, -r\sin \theta} \right]} \\ y & =r\sin \theta \implies \operatorname{grad}_{r, \theta} y = {\left[ {\sin \theta, r\cos \theta} \right]} .\end{align*} Then \begin{align*} u_r &= u_x x_r + u_y y_r \\ &= u_x \cos \theta + u_y \sin \theta \\ &= v_y \cos \theta - v_x \sin \theta \\ &= r^{-1}\qty{v_y \cdot r\cos\theta - u_y \cdot r \sin \theta} \\ &= r^{-1}\qty{v_y y_\theta + u_y x_\theta} \\ &= r^{-1}v_\theta .\end{align*} Similarly \begin{align*} v_r &= v_x x_r + v_y y_r \\ &= v_x \cos \theta + v_y \sin \theta \\ &= -u_y \cos \theta + u_x \sin \theta \\ &= -r^{-1}\qty{u_y \cdot r\cos\theta i u_x \cdot r\sin \theta } \\ &= -r^{-1}\qty{u_x x_\theta + u_y y_\theta} \\ &= -r^{-1}u_\theta .\end{align*}

Part 2:

Define \(u(r, \theta) = \log(r)\) and \(v(r, \theta) = \theta\) to write \(\operatorname{Log}(z) = u+iv\). Then check \begin{align*} u_r &= r^{-1}, \quad v_\theta = 1 \implies u_r = r^{-1}v_\theta \\ v_r &= 0, \quad u_\theta = 0 \implies v_r = -r^{-1}u_\theta ,\end{align*} provided \(r>0\) so that \(u_r\) is defined.

That this function is not continuous: let \(w_k = 1\cdot e^{i(2\pi - 1/k)}\), noting that these are two sequences converging to 1. If \(\operatorname{Log}(z)\) were continuous, we would have \begin{align*} \lim_{k\to\infty} \operatorname{Log}(w_k) = \operatorname{Log}(1) \coloneqq\log(1) + i\cdot 0 = 0 ,\end{align*} Thus for any \({\varepsilon}\) we could choose \(k\gg 1\) so that \begin{align*} {\left\lvert {\log(z_k) - 0} \right\rvert}, {\left\lvert {\log(w_k) - 0 } \right\rvert} < {\varepsilon} .\end{align*} However, \begin{align*} \log(w_k) = \log(1) + i(2\pi - 1/k) = i(2\pi - 1/k) = 2\pi i - {1\over k} > {\varepsilon} ,\end{align*} for arbitrarily large \(k\), provided we choose \({\varepsilon}\) small.

Let \(\tilde f(z) \coloneqq f(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\). Using that \(f\) is analytic iff its components solve Cauchy-Riemann, using that \(f, \tilde f\) are analytic, \begin{align*} u_x = v_y && u_y = -v_x \\ u_x = -v_y && u_y = v_x \\ \\ \implies 2u_x = v_y - v_y = 0 \implies u_x = 0 \\ \implies 2u_y = v_x - v_x = 0 \implies u_y = 0 \\ \implies 0 = u_y - u_y = v_x - (-v_x) = 2v_x \implies v_x = 0 \\ \implies 0 = u_x - u_x = v_y - (-v_y) = 2v_y \implies v_y = 0 ,\end{align*} so \(\operatorname{grad}u = [u_x, u_y] \equiv \mathbf{0}\) making \(u\) constant. Similarly \(\operatorname{grad}v = [v_x, v_y] = \mathbf{0}\), so \(f: {\mathbb{R}}^2\to {\mathbb{R}}\) is constant.

Suppose \(fg=0\) on \(D\) but \(f\not\equiv 0\), we’ll show \(g\equiv 0\) on \(D\). Since \(f\not \equiv 0\), \(f(z_0)\neq 0\) at some point \(z_0\). Since \(f\) is holomorphic, in particular \(f\) is continuous, so there is a neighborhood \(U\ni z_0\) where \(f(z)\neq 0\) for any \(z\in U\). But \(f(z)g(z) = 0\) for all \(z\in U\), and since \({\mathbb{C}}\) is an integral domain, this forces \(g(z) = 0\) for every \(z\in U\). So \(g\equiv 0\) on \(U\). Now \(U\) is a set with a limit point, so by the identity principle, \(g\equiv 0\) on \(D\).

Part a: Not possible: if \(f\) is holomorphic then \(f\) is in particular continuous, so \begin{align*} f(0) = f(\lim 1/n) = \lim f(1/n) = \lim (-1)^n ,\end{align*} which does not converge.

Part b: Not possible: note that \(1/n\) has a limit point, so if \(f(1/n)=0\) then \(f\equiv 0\) on \({\mathbb{D}}\) by the identity principle. In particular, we can not have \(f(1/n) = e^{-n}>0\).

Alternatively, note that a holomorphic \(f\) must have isolated zeros, while \(z_0=0\) is forced to be a zero of \(f\) by continuity, which has infinitely many zeros of the form \(1/n\) in any neighborhood.

Part c: Not possible: suppose so, then by continuity, we have \begin{align*} f(0) = f(\lim 1/n^2)= \lim f(1/n^2)=\lim 1/n = 0 ,\end{align*} so \(z_0=0\) is a zero. Now defining \(g(z) = z^{1\over 2} \coloneqq e^{1\over 2 \log(z)}\) on \(U \coloneqq{\mathbb{C}}\setminus(-\infty, 0]\) extending this continuously to zero by \(g(0)= 0\) yields \(g(z) = f(z)\)on \(\left\{{1/n^2 {~\mathrel{\Big\vert}~}n>1}\right\}\cup\left\{{0}\right\}\), so \(g(z) \equiv f(z)\) on \(U\). But then \(g\equiv f\) on \({\mathbb{D}}\), and \(g\) is not holomorphic on all of \({ \mathsf{D} }\), contradicting that \(f\) was holomorphic on \({\mathbb{D}}\).

Part d: Yes: note that this forces \(f(0) = \lim {n-2\over n-1} = 1\) by continuity at \(z=0\). We can write \begin{align*} {n-2\over n-1} = {1 - 2\cdot{1\over n} \over 1 - {1\over n}} ,\end{align*} so define \(g(z) \coloneqq{1-2z\over 1-z}\). Then \(g(1/n) = f(1/n)\) for all \(n\) and \(g(0) = 1= f(0)\), so \(g=f\) on a set with an accumulation point making \(g\equiv f\) on \({\mathbb{D}}\). Note that \(g\) is holomorphic on \({\mathbb{D}}\), since it has only a simple pole at \(z_0 = 1\).

\begin{align*} {\left\lvert {w_1 - c^2 w_2} \right\rvert}^2 &= (w_1 - c^2 w_2) ( \mkern 1.5mu\overline{\mkern-1.5muw_1\mkern-1.5mu}\mkern 1.5mu - c^2 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu ) \\ &= {\left\lvert {w_1} \right\rvert}^2 + c^4 {\left\lvert {w_2} \right\rvert}^2 - 2c^2 \Re(w_1 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu) \\ &= {\color{green} c^2 {\left\lvert {w_2} \right\rvert}^2 } + c^4 {\left\lvert {w_2} \right\rvert}^2 - 2c^2 \Re(w_1 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu) \\ &= c^2 {\left\lvert {w_2} \right\rvert}^2 + {\color{green} c^2 {\left\lvert {w_1} \right\rvert}^2 } - 2c^2 \Re(w_1 \mkern 1.5mu\overline{\mkern-1.5muw_2\mkern-1.5mu}\mkern 1.5mu) \\ &= c^2 {\left\lvert {w_1 - w_2} \right\rvert} ,\end{align*} where we’ve applied the assumption \({\left\lvert {w_1} \right\rvert} = c{\left\lvert {w_2} \right\rvert}\) twice.

Using part 1: \begin{align*} w_1\coloneqq z-z_1, w_2 \coloneqq z-z_2 \implies {\left\lvert {w_1} \right\rvert} &= c{\left\lvert {w_2} \right\rvert} \\ \implies {\left\lvert {w_1 - c^2 w_2} \right\rvert} &= c {\left\lvert {w_1 - w_2} \right\rvert} \\ \implies {\left\lvert { z-z_1 - c^2 (z-z_2) } \right\rvert} &= {\left\lvert {(z-z_1) - (z-z_2)} \right\rvert} \\ \implies {\left\lvert {(1-c^2) z - z_3} \right\rvert} &= {\left\lvert { z_2 - z_1 } \right\rvert} \\ \implies {\left\lvert {z-z_4} \right\rvert} &= r ,\end{align*} where the \(z_i\) and \(r\) are all constant, so this is the equation of a circle.

• A circle of radius 1 about \(z=1\).

• A circle, using that Apollonius circles are characterized as the locus of distances whose ratios to some fixed points \(A, B\) are constant. To actually compute this: \begin{align*} {\left\lvert {z-1} \right\rvert}^2 &= 4{\left\lvert {z-2} \right\rvert}^2 \\ \implies (x-1)^2 + y^2 - 4\qty{(x-2)^2 + y^2 } &=0 \\ -3x^2 + 14x - 3y^2 - 15 &= 0 && \star\\ \implies x^2 - {14\over 3}x + y^2 + 5 &= 0 \\ \implies (x- {14\over 2\cdot 3})^2 - {14\over 2\cdot 3}^2 + y^2 + 5 &= 0 \\ \implies (x-{14\over 6})^2 + y^2 = \qty{2\over 3}^2 ,\end{align*} which is a circle of radius \(2/3\) with center \(\qty{{14\over 6}, 0}\). To avoid the calculation, use \begin{align*} Ax^2 + Bxy + Cy + \cdots = 0,\quad A=1, B=0, C=1 \implies \Delta \coloneqq B^2 - 4AC < 0 ,\end{align*} which is an ellipse, and since \(A=C\) it is in fact a circle.

• \(S^1\), using that \({1\over z} = {\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \over z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} = {\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\over {\left\lvert {z} \right\rvert}^2}\) and if this equals \(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\), then \({\left\lvert {z} \right\rvert}^2=1\). Alternatively, \(1 = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5muz = {\left\lvert {z} \right\rvert}^2\).

• Vertical line through \(z=3\).

• Horizontal line through \(z=ia\).

• Region to the right of the vertical line through \(z=a\).

• Exterior of a circle: same calculation is (2), replacing \(=0\) with \(<0\). Note that the line marked \(\star\) involves dividing by a negative, so this flips the sign, and we get \(\cdots > \qty{2\over 3}^2\) at the end.

The geometry at hand:

By the inscribed angle theorem, this locus is an arc of a circle whose center \(O\) is the point for which the angle \(aOb\) is \(2\theta\):

\(\implies\): Write the vertices as \(z_1, z_2, z_3\) and the sides as

• \(s_1 \coloneqq z_2-z_1\)
• \(s_2 \coloneqq z_3 - z_2\)
• \(s_1 \coloneqq z_1 -z_3\)

Note that \(s_i = \pm \zeta_3 s_{i-1}\), dividing yields \begin{align*} {s_2 \over s_3} &= {s_1\over s_2} \\ &\iff s_2^2 - s_1 s_3 = 0 \\ &\iff \left(z_{2}-z_{3}\right)^{2}-\left(z_{2}-z_{1}\right)\left(z_{1}-z_{3}\right)=0 \\ &\iff \left(z_{2}^{2}+z_{3}^{2}-2 z_{2} z_{3}\right)-\left(z_{2} z_{1}-z_{2} z_{3}-z_{1}^{2}+z_{1} z_{3}\right)=0 \\ &\iff z_{1}^{2}+z_{2}^{2}+z_{3}^{2}-\left(z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}\right)=0 .\end{align*}

\(\impliedby\): We still have \(s_i = \theta_i s_{i-1}\) for some angles \(\theta_i\) We have

and \begin{align*} {s_1\over s_2} &= {\theta_1 \over \theta_2} \cdot {s_3\over s_1} \\ {s_2\over s_3} &= {\theta_2 \over \theta_3} \cdot {s_1\over s_2} \\ {s_3\over s_1} &= {\theta_3 \over \theta_1} \cdot {s_2\over s_3} .\end{align*}

Running the above calculation backward yields \(s_2/s_3 = s_1/s_2\), and by the 2nd equality above, this forces \(\theta_2 = \theta_3\). Similar arguments show \(\theta_1=\theta_2 = \theta_3\) which forces \(s_1=s_2 = s_3\).

First note that by the maximum modulus principal, it suffices to consider sups on the boundary, i.e. \begin{align*} M_R = \sup_{{\left\lvert {z} \right\rvert} = R}{\left\lvert {f(z)} \right\rvert}, \qquad N_R = \sup_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {f'(z)} \right\rvert} .\end{align*}

The first estimate: stuck!

The second estimate: suppose \(z_0 \in {\mathbb{D}}_R(0)\), then any \(D_R(z_0)\) is contained in \(D_{2R}(0)\), So for any such \(z_0\), apply Cauchy’s integral formula centered at \(z_0\): \begin{align*} f^{(1)}(z_0) &= {1\over 2\pi i }\oint_{{{\partial}}{\mathbb{D}}_{R}(z_0)} {f(\xi)\over (\xi-z_0)^2 }\,d\xi\\ \implies {\left\lvert { f^{(1)}(z_0)} \right\rvert} &\leq {1\over 2\pi} \oint_{{{\partial}}{\mathbb{D}}_{R}(z_0)} {\left\lvert {f(\xi)\over (\xi-z_0)^2 } \right\rvert}\,d\xi\\ &= {1\over 2\pi} \oint_{{{\partial}}{\mathbb{D}}_R(z_0)} { {\left\lvert {f(\xi)} \right\rvert} \over {\left\lvert {\xi-z_0} \right\rvert}^2 } \,d\xi\\ &= {1\over 2\pi} \oint_{{{\partial}}{\mathbb{D}}_R(z_0)} { {\left\lvert {f(\xi)} \right\rvert} \over R^2 } \,d\xi\\ &\leq {1\over 2\pi} R^{-2} \oint_{{{\partial}}{\mathbb{D}}_R(z_0)} { \sup_{z\in {{\partial}}{\mathbb{D}}_{R}(z_0) } {\left\lvert {f(z)} \right\rvert} } \,d\xi\\ &= {1\over 2\pi} R^{-2} \sup_{{{\partial}}{\mathbb{D}}_R(z_0) } {\left\lvert {f(z)} \right\rvert} \cdot 2\pi R \\ &= R^{-1}\sup_{{{\partial}}{\mathbb{D}}_R(z_0) } {\left\lvert {f(z)} \right\rvert} \\ &\leq R^{-1}M_{2R} ,\end{align*} where we’ve used in the last step that \({\mathbb{D}}_R(z_0) \subseteq {\mathbb{D}}_{2R}(0)\), and sups can only get larger when taken over larger sets. Since this was an arbitrary \(z_0\in {\mathbb{D}}_R(0)\), this holds for all \(z\) with \({\left\lvert {z} \right\rvert} \leq R\). Since taking sups preserves inequalities, we have \begin{align*} {\left\lvert {f'(z_0)} \right\rvert} \leq R^{-1}M_{2R}\, \forall {\left\lvert {z} \right\rvert} \leq R \implies N_R\coloneqq\sup_{{\left\lvert {z} \right\rvert} \leq R}{\left\lvert {f'(z)} \right\rvert} \leq R^{-1}M_{2R} .\end{align*}

Part 1: For ease of notation, let \(z=z_1\) and \(w=z_2\) We want to show \begin{align*} {\left\lvert {1- z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \right\rvert} - {\left\lvert {z-w} \right\rvert}^2 = \qty{{1 - {\left\lvert {z} \right\rvert}^2 }} \qty{{1 - {\left\lvert {w} \right\rvert}^2}} .\end{align*}

So write \begin{align*} {\left\lvert {1- z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \right\rvert} - {\left\lvert {z-w} \right\rvert}^2 &= (1-z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu)(1-\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w) - (z-w)(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu) \\ &= 1 - z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w - {\left\lvert {z} \right\rvert}^2{\left\lvert {w} \right\rvert}^2 - {\left\lvert {z} \right\rvert}^2 - {\left\lvert {w} \right\rvert}^2 + w\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu \\ &= 1 - {\left\lvert {z} \right\rvert}^2{\left\lvert {w} \right\rvert}^2 - {\left\lvert {z} \right\rvert}^2 - {\left\lvert {w} \right\rvert}^2 \\ &=(1-{\left\lvert {z} \right\rvert}^2)(1-{\left\lvert {w} \right\rvert}^2) .\end{align*}

Part 2 and 3: \begin{align*} {\left\lvert {z-w \over 1 - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w} \right\rvert}^2 \leq 1 &\iff 0 \leq {\left\lvert {1 - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w} \right\rvert}^2 - {\left\lvert {z-w} \right\rvert}^2 \\ &\iff 0 \leq (1 - {\left\lvert {z} \right\rvert}^2 )(1 - {\left\lvert {w} \right\rvert}^2) ,\end{align*} where we’ve used part 1. But this is clearly true when \({\left\lvert {z} \right\rvert}, {\left\lvert {w} \right\rvert} < 1\), so the RHS is positive. Moreover if \({\left\lvert {z} \right\rvert} = {\left\lvert {w} \right\rvert} = 1\), the RHS is zero, yielding equalities at every step instead.

Part 1: See Spring 2021.1 above.

Part 2, holomorphicity: This is clearly meromorphic, as it’s a rational function, and has a singularity only at \(z\) such that \(\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z = 1\). This can only happen if \(z, w \in S^1\): taking the modulus yields \begin{align*} \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z = 1 \implies {\left\lvert {w} \right\rvert}^2{\left\lvert {z} \right\rvert}^2 = 1 ,\end{align*} and moreover since \({\left\lvert {w} \right\rvert}^2 \leq 1\) and \({\left\lvert {z} \right\rvert}^2\leq 1\), the only way this product can be one is when \({\left\lvert {w} \right\rvert}^2 = {\left\lvert {z} \right\rvert}^2 = 1\). This also forces \(z=1/\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu\).

The claim is that the singularity \(1/\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu\) is removable. Note that \(1\over w = \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu\) on the circle, so \(1/\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu = \mkern 1.5mu\overline{\mkern-1.5mu\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu\mkern-1.5mu}\mkern 1.5mu = 2\), so \begin{align*} \qty{ z- \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu^{-1}} \qty{w-z \over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} &= \qty{\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z - 1 \over \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu} \qty{w-z \over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \\ &= \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu^{-1}(w-z) \\ &= w(w-z) \\ &\overset{z\to \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu^{-1}=w }\to 0 .\end{align*}

Part 2, being a bijection: This follows from finding an explicit inverse, using that \(F^2 = \operatorname{id}\): \begin{align*} F(F(z)) &= \frac{w- \qty{ \frac{w-z}{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} } }{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu \qty{ \frac{w-z}{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} } } \\ &= \frac{w(1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z)-(w-z)}{q-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu(w-z)} \\ &= \frac{w-|w|^{2} z-w+z}{1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z-|w|^{2}+\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \\ &= \frac{z\left(1-|w|^{2}\right)}{1-|w|^{2}} \\ &= z .\end{align*}

Part 2, being an involution: A direct check shows that \(F(w) = 0\), since the numerator vanishes, and \(F(0) = {w - 0 \over 1 - 0} = w\).

Part 3, preserving the circle: Follows from the estimate in part 1.

\begin{align*} {\left\lvert {z-w} \right\rvert}^2 &= (z-w)(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu) \\ &= {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - z\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu - \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu w \\ &= {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - 2\Re(\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) \\ &\geq {\left\lvert {z} \right\rvert}^2 + {\left\lvert {w} \right\rvert}^2 - 2{\left\lvert {\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu } \right\rvert}{\left\lvert {z} \right\rvert} \\ &\geq \qty{ {\left\lvert {z} \right\rvert} - {\left\lvert {w} \right\rvert} }^2 ,\end{align*} and taking square roots introduces an absolute value on the final term. Here we’ve used the basic estimate \begin{align*} \Re(z) \leq {\left\lvert {z} \right\rvert} \implies -\Re(z) \geq -{\left\lvert {z} \right\rvert} .\end{align*}

\begin{align*} \prod_{1\leq k \leq n-1} \sin\qty{k\pi\over n} &= \prod_{1\leq k \leq n-1} {\omega_n^k - \omega_n^{-k} \over 2i} \\ &= \qty{1\over 2i}^{n-1} \prod_{1\leq k \leq n-1} \omega_n^{k} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2i}^{n-1} \prod_{1\leq k \leq n-1} \exp\qty{i\pi k\over n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2i}^{n-1} \exp\qty{ {i\pi\over n} \displaystyle\sum_{1\leq k \leq n-1} k } \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2i}^{n-1} e^{i\pi n(n-1)\over 2n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= \qty{1\over 2}^{n-1}\qty{1\over i}^{n-1} \qty{ e^{i\pi \over 2} }^{n-1} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= 2^{1-n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{-k}} \\ &= 2^{1-n} \prod_{1\leq k \leq n-1} \qty{1 - \zeta_n^{k}} \\ &= 2^{1-n} { \Phi_n(z) \over z-1}\Big|_{z=1} \\ &= 2^{1-n} \qty{ \sum_{0\leq k \leq n-1} z^k}\Big|_{z=1} \\ &= n2^{1-n} .\end{align*}

\begin{align*} \prod_{1\leq j\leq n-1} \sin \left(\frac{j \pi}{n}\right) &=\prod_{1\leq j\leq n-1} \frac{1}{2 i} \qty{\zeta_n^{j\over 2} - \zeta_n^{- {j \over 2} }} \\ &=\left(\frac{1}{2 i}\right)^{n-1} \prod_{1\leq j\leq n-1} \zeta_n^{j\over 2} \prod_{1\leq j \leq n-1} \qty{1 - \zeta_n^{-j}} \\ &=\left(\frac{1}{2 i}\right)^{n-1} \prod_{1\leq j\leq n-1} \exp\qty{ij\pi \over n} \prod_{1\leq j \leq n-1} \qty{1 - \zeta_n^{-j}} \\ &=\left(\frac{1}{2 i}\right)^{n-1} \exp \left(\sum_{1\leq j\leq n-1} \frac{i j \pi}{n}\right) \Phi_{n}(1)\\ &=\left(\frac{1}{2 i}\right)^{n-1} \exp \left(\frac{(n-1) i \pi}{2}\right) \Phi_{n}(1)\\ &=\left(\frac{1}{2 i}\right)^{n-1}\left(e^{i \pi / 2}\right)^{n-1} \Phi_{n}(1)\\ &=\left(\frac{1}{2 i}\right)^{n-1} i^{n-1} \Phi_{n}(1)\\ &=\left(\frac{1}{2}\right)^{n-1} \Phi_{n}(1)\\ &=\frac{n}{2^{n-1}} .\end{align*}

\begin{align*} 0 &\leq (1 - {\left\lvert {w} \right\rvert}^2)(1-{\left\lvert {z} \right\rvert}^2) \\ \implies {\left\lvert {w} \right\rvert}^2 + {\left\lvert {z} \right\rvert}^2 &\leq 1 + {\left\lvert {w} \right\rvert}^2 {\left\lvert {z} \right\rvert}^2 \\ \implies {\left\lvert {w} \right\rvert}^2 + {\left\lvert {z} \right\rvert}^2 - 2\Re(\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) &\leq 1 + {\left\lvert {w} \right\rvert}^2 {\left\lvert {z} \right\rvert}^2 - 2\Re(\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) \\ \implies {\left\lvert {w-z} \right\rvert}^2 &\leq {\left\lvert {1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5muz} \right\rvert}^2 .\end{align*} Note that if either \({\left\lvert {w} \right\rvert}^2 = 1\) or \({\left\lvert {z} \right\rvert}^2 = 1\) then the first line is an equality, yielding equality in the final line.

• That \(F: {\mathbb{D}}\to {\mathbb{D}}\): follows from the inequality, since \({\left\lvert {z} \right\rvert}, {\left\lvert {w} \right\rvert} < 1\) for \(z,w\in {\mathbb{D}}\). Holomorphicity: follows from the fact that rational expressions of holomorphic functions are holomorphic away from where the denominators vanish.

• Then just note that \({\left\lvert {\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \right\rvert} \leq {\left\lvert {w} \right\rvert}{\left\lvert {z} \right\rvert} < 1\), so \({\left\lvert {1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \right\rvert} > 0\).

• \(F(0) = {w-0 \over 1-0} = w\)

• \(F(w) = {w-w\over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu w} = 0\)

• \({\left\lvert {z} \right\rvert} = 1\) yields equality in part 1.

Other notes: \(F\) is bijective on \({\mathbb{D}}\): \begin{align*} F(F(z)) &= {w - \qty{w-z\over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu\qty{w-z\over 1 - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} } \\ &= {w(1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) - (w-z) \over (1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z) - \mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu (w-z)} \\ &= {z-{\left\lvert {w} \right\rvert}^2 z \over 1 - {\left\lvert {w} \right\rvert}^2 }\\ &= z .\end{align*}

\hfill

\hfill

Let \(f(z) = {z+1\over z(z-1)}\).

About \(z=0\):

\begin{align*} f(z)  &= (z+1) \qty{- {1 \over z} + {1\over z-1} } \\ &=  -(z+1) \qty{{1\over z} + \sum_{n=0}^\infty z^n } \\ &= -(z+1)\sum_{n=-1}^\infty z^n \\ &= {1\over z} + 2\sum_{n=0}^\infty z^n \\ &= -{1\over z} -2 - 2z - 2z^2 - \cdots .\end{align*}

About \(z=1\):

\begin{align*} f(z)  &= \qty{(1-z) -2 \over 1-z} \qty{1 \over 1 - (1-z)} \\ &= \qty{1 - {2\over 1-z}} \sum_{n=0}^\infty (1-z)^n \\  &= \sum_{n=0}^\infty (1-z)^n - 2 \sum_{n=-1}^\infty (1-z)^n \\ &= -{2\over 1-z} - \sum_{n=0}^\infty (1-z)^n \\ &= {2\over z-1} + \sum_{n=0}^\infty (-1)^{n+1} (z-1)^n \\ &= {2\over z-1} - 1 + (z-1) - (z-1)^2 + \cdots .\end{align*}

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Let \(f(z) = {z+1\over z(z-1)}\).

About \(z=0\):

\begin{align*} f(z)  &= (z+1) \qty{- {1 \over z} + {1\over z-1} } \\ &=  -(z+1) \qty{{1\over z} + \sum_{n=0}^\infty z^n } \\ &= -(z+1)\sum_{n=-1}^\infty z^n \\ &= {1\over z} + 2\sum_{n=0}^\infty z^n \\ &= -{1\over z} -2 - 2z - 2z^2 - \cdots .\end{align*}

About \(z=1\):

\begin{align*} f(z)  &= \qty{(1-z) -2 \over 1-z} \qty{1 \over 1 - (1-z)} \\ &= \qty{1 - {2\over 1-z}} \sum_{n=0}^\infty (1-z)^n \\  &= \sum_{n=0}^\infty (1-z)^n - 2 \sum_{n=-1}^\infty (1-z)^n \\ &= -{2\over 1-z} - \sum_{n=0}^\infty (1-z)^n \\ &= {2\over z-1} + \sum_{n=0}^\infty (-1)^{n+1} (z-1)^n \\ &= {2\over z-1} - 1 + (z-1) - (z-1)^2 + \cdots .\end{align*}

Write \begin{align*} P(z) &= \prod_{k\leq n} (z-a_k) \\ Q(z) &= \prod_{k\leq m}(z-b_k) \\ Q_j(z) &= \prod_{k\neq j}(z-b_k) = {Q(z) \over z-z_j} .\end{align*}

For \(b_\ell\) a simple pole, \begin{align*} {P(z) \over Q(z) } = {1\over z-b_\ell} {P(z) \over Q_\ell(z)} &= {1\over z-b_\ell}\qty{c_0 + c_1(z-b_\ell) + c_2(z-b_\ell)^2 + \cdots} \\ &= {c_0 \over z-b_\ell} + c_1 + { \mathsf{O}} (z-b_\ell) \\ & \coloneqq{P_{b_\ell}(z)} + c_1 + { \mathsf{O}} (z-b_\ell) ,\end{align*} so the principal part at \(z=z_\ell\) is given by \begin{align*} P_{z_\ell}(z) = {c_0 \over z-b_\ell} = {P(z) \over Q_\ell(z)}\Big|_{z=b_\ell} = \lim_{z\to z_\ell} {(z-b_\ell) P(z) \over Q(z)} .\end{align*}

For \(b_\ell\) a double pole, \begin{align*} {P(z) \over Q(z) } &= {1\over (z-b_\ell)^2 } {(z-z_\ell)^2P(z) \over Q(z) } \\ &= {1\over (z-b_\ell)^2}\qty{ d_0 + d_1(z-b_\ell) + d_2(z-b_\ell)^2 } \\ &= {d_0 \over (z-b_\ell)^2} + {d_1\over z-z_\ell} + d_2 + { \mathsf{O}} (z-b_\ell) \\ &\coloneqq P_{b_\ell}(z) + d_2 + { \mathsf{O}} (z-b_\ell) .\end{align*} To extract the \(d_1\) coefficient, note that \begin{align*} {(z-b_\ell)^2 P(z) \over Q(z)} &= d_0 + d_1(z-b_\ell) + \cdots \\ \implies {\frac{\partial }{\partial z}\,} {(z-b_\ell)^2 P(z) \over Q(z)} &= d_1 + 2d_2(z-b_\ell) + \cdots ,\end{align*} so \begin{align*} d_0 &= \lim_{z\to b_\ell} { (z-b_\ell)^2 P(z) \over Q(z) } \\ d_1 &= \lim_{z\to b_\ell} {\frac{\partial }{\partial z}\,} {(z-b_\ell)^2 P(z) \over Q(z) } \\ P_{b_\ell} &= {d_0 \over (z-b_\ell)^2} + {d_1\over z-b_\ell} .\end{align*}

Compute the series expansion of the RHS: \begin{align*} (z-1) f(z) &= (z-1) \sum_{n\geq 1} c_n z^k \\ &= -c_1z + \sum_{n\geq 2} (c_{n-1} - c_n) z^n \\ &\overset{z\to 1}\longrightarrow -c_1 + \sum_{n\geq 2} c_{n-1} - c_n \\ &\coloneqq\lim_{N\to\infty} -c_1 z + \sum_{n=2}^N c_{n-1} - c_n \\ &= \lim_{N\to\infty} -c_N ,\end{align*} where we’ve used that the sum is telescoping.

If \(f\) is entire, write \(f(z) = \sum_{k\geq 0}c_k z^k\) and \(g(z) \coloneqq f(1/z) = \sum_{k\geq 0}c_k z^{-k}\). If \(z=\infty\) is a pole of order \(m\) of \(f\), \(z=0\) is a pole of order \(m\) of \(g\), so \begin{align*} g(z) = \sum_{0\leq k \leq m}c_k z^{-k} \implies f(z) = \sum_{0\leq k \leq m}c_k z^k ,\end{align*} making \(f\) a polynomial of degree at most \(m\).

Write \(f(z) = P_2(z) + g(z)\) where \(P_2\) is the principal part of \(f\) at \(z=2\) and \(g\) is holomorphic at \(z=2\). Then \(g\) is an entire function with a double pole at \(\infty\), and is thus a polynomial of degree at most \(2\), so \(g(z) = c_2z^2 + c_1 z + c_0\). Since the pole of \(f\) at \(z=2\) is simple, \(P_2(z) = \sum_{k\geq -1} d_k (z-2)^k\). Combining these, we can write \begin{align*} f(z) = d_{-1}(z-2)^{-1}+ \sum_{0\leq k\leq 3} (d_k + c_k)(z-2)^k + \sum_{k\geq 3}d_k (z-2)^k .\end{align*} However, if \(d_k\neq 0\) for any \(k\geq 3\), this results in a higher order pole at \(\infty\), so \(f\) must be of the form \begin{align*} f(z) = d_{-1}(z-2)^{-1}+ \sum_{0\leq k\leq 3} (d_k + c_k)(z-2)^k .\end{align*}

Write \(f(z) = \sum_{k\geq 0} c_k z^k\) since it is entire.

• If \(f\) has finitely many zeros, \(f\) is nonconstant and entire, and thus unbounded by Liouville. If \(f\) is nonconstant, \(z=\infty\) can not be removable, since this would force \(f\) to be constant. So \(z=\infty\) can be a pole or an essential singularity. Both possibilities can occur: if \(f\) is a polynomial, it is entire with finitely many zeros and a pole at \(z=\infty\). Taking \(f(z)= e^z\) has no zeros and an essential singularity at \(z=\infty\).

• If \(f\) has infinitely many zeros, if \(f\) is nonconstant then infinitely many \(c_k\) are nonzero – otherwise \(f\) is a polynomial and can only have finitely many zeros. Then \(g(z) \coloneqq f(1/z) = \sum_{k\geq 0}{c_k\over z^k}\) has infinitely many nonzero terms, making \(z=0\) an essential singularity for \(g\) and \(z=\infty\) essential for \(f\).

Part 1: This is clear: \(\sin^2(\pi z) = 0 \iff z = k\) for \(k\in {\mathbb{Z}}\), and this is a pole of order 2 for \(f\). Every \(k\in {\mathbb{Z}}\) is visibly an order 2 pole of \(g\).

Part 2: By periodicity, it suffices to consider the singularity at \(z_0 = 0\). Expanding \(\sin(\pi z) = \pi z - {1\over 3!}(\pi z)^3 + {1\over 5!} (\pi z)^5 + \cdots\) and considering \(\sin(\pi z)^2\) shows that \(z=0\) is a pole of order 2. So \(z^2f(z)\) has a removable singularity at \(z=0\), and can be expanded: \begin{align*} z^2f(z) &= \qty{\pi z\over \sin(\pi z)}^2 \\ &= (\pi z)^2 \qty{ (\pi z) ^{-1}+ {1\over 3!}(\pi z) + {7\over 360} (\pi z^3) + \cdots}^2 \\ &= (\pi z)^2 \qty{ (\pi z)^{-2} + { \mathsf{O}} (1) } \\ &= 1 + { \mathsf{O}} (z^2) \\ \implies f(z) &= z^{-2} + { \mathsf{O}} (1) ,\end{align*} so the singular part of \(f\) at \(z=0\) is \(z^{-2}\). This coincides with the \({1\over z^2}\) term in \(g\). The remaining principal parts at \(z=k\) are \({1\over (z-k)^2},\) using the fact that \(f(z+1) = f(z)\), so \(f(k) = f(0)\) and the Laurent expansions are gotten by substituting \(z-k\) in for \(z\) everywhere.

Part 3: Periodicity is clear for \(f\). For \(g\), \begin{align*} g(z+1) = \sum_{k\in {\mathbb{Z}}} ((z-1)-k)^{-2} = \sum_{k'\in {\mathbb{Z}}} (z-k)^{-2} ,\end{align*} where \(k' \coloneqq k+1\), and the equality is true since both sums run over all of \({\mathbb{Z}}\).

For convergence: take \(z=it\), then for \(f\) \begin{align*} f(it) \sim \csc^2(i\pi t) &\sim \qty{ e^{i\pi (it) } - e^{-i\pi (it)}}^{-2} \\ &= \qty{e^{-\pi t} - e^{\pi t}}^{-2} \\ &\leq {1\over e^{-\pi t} + e^{\pi t} } \\ &\sim e^{-\pi t} \\ &\to 0 ,\end{align*} using the reverse triangle inequality and that the \(e^{-\pi t}\) term in the denominator is negligible for large \(t\).

For \(g\), \begin{align*} g(it) &\sim t^{-2} + \sum_{k\geq 1} (t^2 + k^2)^{-1}\\ &\leq t^{-2} + \sum_{1\leq k \leq N}(t^2 + k^2)^{-1}+ \sum_{k\geq N}(t^2 + k^2)^{-1}\\ &\leq t^{-2} + \sum_{1\leq k \leq N}(t\cdot k^2)^{-1}+ \sum_{k\geq N}(k^2)^{-1}\\ &\leq t^{-2} + t^{-1}\sum_{1\leq k \leq N}(k^2)^{-1}+ \sum_{k\geq N}(k^2)^{-1}\\ &\overset{N\to\infty\implies t\to\infty}\longrightarrow 0 ,\end{align*} where given \(N\) we can pick \(t\) large enough so that \(t^2 + k^2 \geq tk^2\) for all \(k\leq N\). These converge to zero as \(N\to\infty\) since \(\sum k^{-2} < \infty\), making the last term the tail of a convergent sum.

Part 4: Since \(f,g\) uniformly converge to zero on the strip \(0<\Re(x) < 1\), they are bounded on this strip. Since this is a fundamental domain for their periods, they are bounded on \({\mathbb{C}}\). Write \(h\coloneqq f-g\), then \(h\) is entire since \(f,g\) have the same singular parts, and bounded since \({\left\lvert {h} \right\rvert}\leq {\left\lvert {f} \right\rvert} + {\left\lvert {g} \right\rvert}\). By Liouville, \(h\) is constant with \(\lim_{t\to\infty} h(it) = 0\), so \(h\equiv 0\) and \(f\equiv g\).

Note that \(z=0\) can not be a removable singularity, since then \(f\) would extend to a holomorphic function over \(z=0\), and by continuity \(0 = \lim f(z_n) = f(\lim z_n) = f(0)\). By the identity principle, this would force \(f\equiv 0\), contradicting that \(f\) is nonconstant.

It can not be a pole, because then \(f(z_n)\to \infty\), but \({\left\lvert {f(z_n)} \right\rvert} = 0 < {\varepsilon}\) for any \({\varepsilon}\) infinitely many times.

By the ML estimate, \(\int_{C_R} f \to 0\).

The residue contribution: note the simple pole at \(\omega_n \coloneqq e^{i\pi \over n}\), \begin{align*} \mathop{\mathrm{Res}}_{z=\omega_n} f(z) = {1\over n\omega_n^{n-1}} = {\omega \over n\omega^n} = -{\omega_n \over n} .\end{align*}

The segment contributions: \(\int_{\gamma_1}f\to I\), and \begin{align*} \int_{\gamma_2}f(z) \,dz= \int_\infty^0 {1\over 1 + (\zeta_nt)^n} \zeta_n \,dt= -\zeta_n I ,\end{align*} so the contour contributions sum to \((1-\zeta_n)I\).

Solving: \begin{align*} I = -{2\pi i \omega_n\over n(1-\zeta_n)} = -2\pi i {1\over \omega_n^{-1}- \omega_n} = -2\pi i {1\over 2i \sin\qty{\pi \over n}} = {\pi \over n} \csc\qty{\pi \over n} .\end{align*}

Write \(\omega_{n, k} = \exp\qty{(2k+1)i\pi \over n}\) and factor \(z^n+1\) as \begin{align*} z^n+1 = \prod_{1\leq k \leq n}(z-\omega_{n, k}) = (z-e^{i\pi \over n})(z-e^{3i\pi \over n}) \cdots (z-e^{(2n-1)i\pi \over n}) .\end{align*} Note that only the root \(e^{i\pi\over n}\) lies in the \(2\pi/n\) wedge, so it is the only (simple) pole of \(f(z) \coloneqq{1\over 1+z^n}\) in this region. Since the pole is simple, we can compute the residue easily. Write \(r_0 \coloneqq e^{e\pi\over n}\), then By L’Hopital, \begin{align*} \mathop{\mathrm{Res}}_{z = r_0} {1\over 1+z^n} &= \lim_{z\to r_0} {z-r_0 \over 1 + z^n} \\ &= \lim_{z\to r_0} {1\over nz^{n-1}} \\ &= {1\over nr_0^{n-1}} \\ &= {1\over n e^{i\pi (n-1) \over n}} \\ &= n^{-1}{\exp\qty{-i\pi (n-1)\over n }} .\end{align*}

Take a contour \(\Gamma\) comprised of

• \(\gamma_1 = [0, R] \subseteq {\mathbb{R}}\)
• \(\gamma_2 = \left\{{Re^{it} {~\mathrel{\Big\vert}~}t\in [0, 2\pi/n]}\right\}\)
• \(\gamma_3 = \zeta_n [0, R]\)

By the residue theorem \begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=r_0} f(z) = I \coloneqq\int_\Gamma f = \qty{\int_{\gamma_1} + \int_{\gamma_2} + \int_{\gamma_3}}f .\end{align*}

so in the limit we have \begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=r_0}f(z) &= \qty{1 - \zeta_n}\int_{\gamma_1}f \\ \implies \int_{\gamma_1} f &= {2\pi i \mathop{\mathrm{Res}}_{z=r_0}f(z) \over 1 - \zeta_n}\\ &= {2\pi i e^{-\pi (n-1) \over n} \over n\qty{1-e^{2\pi i \over n}}} \\ &= {2\pi i \over n} \left[ e^{i\pi} e^{-i\pi \over n}\qty{1 - e^{2\pi i \over n}} \right]^{-1}\\ &= {2\pi i \over n} \left[ -1\qty{e^{-i\pi \over n} - e^{\pi i \over n}} \right]^{-1}\\ &= {2\pi i \over n} \left[ 2i \sin\qty{\pi\over n} \right]^{-1}\\ &= {\pi \over n\sin\qty{\pi \over n}} .\end{align*}

Sketch: see here.

Write \(I\) for the integral, \(\zeta_3\coloneqq e^{2\pi i\over 3}, \omega_3 \coloneqq e^{i\pi\pver 3}\). Take a indented semicircular wedge \(\Gamma\) at an angle of \(2\pi/3\), noting the pole at \(\omega_3 \coloneqq e^{i \pi \over 3}\):

Choosing a branch cut of \(\log\) along \(\theta = -\pi/2\), so \(\arg(z) \in (-\pi/2, 3\pi/2)\), this makes \(f(z) \coloneqq z^{\alpha-1}/(1+z^3)\) meromorphic on \(\Gamma\).

By the ML estimate, the integrals along \(C_{\varepsilon}, C_R\) will vanish in the limit.

The contribution from the contour: parameterize \(\gamma_2\) as \(\left\{{\zeta_3 t{~\mathrel{\Big\vert}~}t\in [{\varepsilon}, R]}\right\}\), then \begin{align*} \int_{\gamma_2}f(z) \,dz &= \int_R^{\varepsilon}{(\zeta_3 t)^{\alpha - 1} \over 1 + (\zeta_3 t)^3} \zeta_3\,dt\\ &= \zeta_3 \zeta_3^{\alpha-1}\int_R^{\varepsilon}{t^{\alpha-1} \over 1 + t^3}\,dt\\ &\to -\zeta_3^\alpha I ,\end{align*} so the total contribution is \begin{align*} \oint_\Gamma f \to \qty{\int_{\gamma_1} + \int_{\gamma_2}}f = (1-\zeta_3^{\alpha}) I .\end{align*}

Computing the contributions from residues: \begin{align*} \mathop{\mathrm{Res}}_{z=\omega_3} f(z) &= \lim_{z\to \omega_3} {(z-\omega_3) z^{\alpha-1} \over 1+z^3} \\ &{ \overset{\scriptscriptstyle\text{LH}}{=} }\lim_{z\to\omega_3} {z^{\alpha-1} \over 3z^2} \\ &= {1\over 3} \omega_3^{\alpha-3} \\ &= -{1\over 3}\omega_3^\alpha .\end{align*}

Combining it all using the residue theorem: \begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=\omega_3} f(z) &= \oint_\gamma f \\ \implies I &= {-2\pi i\over 3} {\omega_3^\alpha \over 1-\zeta_3^\alpha } \\ &= {-2\pi i\over 3} {1 \over \omega_3^{-\alpha} - \omega_3^\alpha } \\ &= {-2\pi i\over 3} {1 \over -2i\sin\qty{\pi\alpha\over 3} } \\ &= {\pi\over 3}\csc\qty{\alpha\pi\over 3} .\end{align*}

Sketch:

• Set \(z=e^{i\theta}\) to get \begin{align*} \frac{2}{i} \int_{|z|=1} \frac{\mathrm{d} z}{z^{2}+2 a z+1} .\end{align*}

• Factor into two roots \(r_1, r_2\). Use that without loss of generality, \(r_1\in {\mathbb{D}}\) and \(r_2\in {\mathbb{D}}^c\), with neither on \(S^1\) to compute the residue \(4\pi/(r_1 -r_2)\)

Note \(\cosh(z) \coloneqq{1\over 2}(e^z + e^{-z})\), and \begin{align*} \cosh(z) &= 0 \\ \iff e^z + e^{-z} &= 0 \\ \iff e^{-z}(e^{2z} + 1) &= 0 \\ \iff e^{2z} &= -1 \quad \text{since }{\left\lvert {e^{-z}} \right\rvert} = e^{\Re(z)} > 0 \\ \iff 2z &= (2k+1)i\pi \\ \iff z &\in \left\{{\cdots, {-3i\pi \over 2}, {-i\pi \over 2}, {i\pi \over 2}, {3i\pi \over 2}, \cdots}\right\} .\end{align*} So take the following rectangular contour enclosing the singularity \(z= i\pi/2\):

Then letting \(\Gamma\) be the entire contour and \(I\) be the desired integral, we can solve for \(I\): \begin{align*} \int_\Gamma f &= \int_{-R}^R f + \int_{\gamma_{R_1}} f + \int_{\gamma_2}f + \int_{\gamma_{R_2}} f \\ \int_\Gamma f &= I + \int_{\gamma_{R_1}} f + \int_{\gamma_2}f + \int_{\gamma_{R_2}} f \\ \\ I &= \int_{\gamma_{R_1}} f + \int_{\gamma_2}f + \int_{\gamma_{R_2}} f - \int_\Gamma f ,\end{align*} being very sloppy about the fact that we’re going to take \(R\to \infty\).

Computing the residue term \(\int_\Gamma f = 2\pi i \mathop{\mathrm{Res}}_{z=i\pi/2} f(z)\): \begin{align*} {1\over 2\ pi i } \int_\Gamma f &= \mathop{\mathrm{Res}}_{z = i\pi/2} f(z)\\ &= \mathop{\mathrm{Res}}_{z = i\pi/2} {e^{i\xi z} \over \cosh(z) } \\ &= {e^{i\xi z} \over {\frac{\partial }{\partial t}\,} \cosh(z) } \Big|_{i\pi/2}\\ &= {e^{i\xi \cdot i\pi/2} \over \sinh(i\pi/2)} \\ &= {e^{-\xi \pi/ 2} \over i} \\ \implies 2\pi i \mathop{\mathrm{Res}}_{z=i\pi/2} f(z) &= 2\pi e^{-\xi \pi/ 2} .\end{align*} using that \(2\sinh(i\pi/2) = e^{i\pi/2} - e^{-i\pi/2} = i-(-i) = 2i\).

The \(\gamma_2\) term: parameterize \(\gamma_2 = \left\{{t + i\pi {~\mathrel{\Big\vert}~}t\in [-R, R]}\right\}\), then \begin{align*} \int_{\gamma_2} f &= -\int_{-\gamma_2} f \\ &= -\int_{-R}^{R} { e^{iz} \over \cosh(t + i\pi) } \,dz, \quad z = t+i\pi, \,dz= \,dt\\ &= -\int_{-R}^{R} { e^{i\xi(t+i\pi)} \over \cosh(t + i\pi) }\,dt\\ &= -e^{-\xi \pi} \int_{-R}^R {e^{i\xi t} \over \cosh(t+i\pi)} \,dt\\ &= e^{-\xi \pi} \int_{-R}^R {e^{i\xi t} \over \cosh(t)} \,dt\\ &\coloneqq e^{-\xi \pi} I ,\end{align*} using that \(\cosh(z + i\pi) = -\cosh(z)\).

The two \(\gamma_{R_i}\) terms: the claim is that these vanish in the limit \(R\to \infty\). Parameterize \(\gamma_{R_2} = \left\{{R = i \pi t {~\mathrel{\Big\vert}~}t\in [0, 1]}\right\}\), then \begin{align*} {\left\lvert {\int_{\gamma_{R_2}} f} \right\rvert} &= {\left\lvert { \int_0^1 {e^{i\xi(R+i\pi t)} \over \cosh(R + i\pi t)} \,dt} \right\rvert} \\ &= {\left\lvert { 2e^{i\xi R} \int_0^1 {e^{-\xi \pi t } \over {e^{R+ i\pi t} + e^{-R - i\pi t}} } \,dt} \right\rvert} \\ &\leq 2 \int_0^1 {\left\lvert { {e^{-\xi \pi t } \over {e^{R+ i\pi t} + e^{-R - i\pi t}} } } \right\rvert} \,dt\\ &\leq 2 \int_0^1 {e^{-\xi \pi t } \over { {\left\lvert { e^{R+ i\pi t} } \right\rvert} - {\left\lvert { e^{-R - i\pi t} } \right\rvert} } }\,dt\\ &= 2 \int_0^1 {e^{-\xi\pi t} \over e^R - e^{-R} } \\ &= { 2\over e^R - e^{-R}} \int_0^1 e^{-\xi \pi t} \,dt\\ &= {2\over e^R - e^{-R}} \qty{-1\over \xi \pi} e^{\xi \pi t}\Big|_{t=0}^{t=1} \\ &= {2\over e^R - e^{-R}} \qty{1 - e^{-\pi \xi } \over \xi \pi} \\ &\overset{R\to\infty}\longrightarrow 0 .\end{align*}

Putting it all together: \begin{align*} (1 + e^{-\xi \pi}) I &= 2\pi e^{-\xi \pi / 2} \\ \implies I &= {2\pi e^{-\xi \pi / 2} \over 1 + e^{-\xi \pi }} \\ &= {2\pi \over e^{\xi\pi/2} (1 + e^{-\xi \pi })} \\ &= {2\pi \over e^{\xi\pi/2} + e^{-\xi\pi/2} } \\ &= {2\pi \over 2\cosh(\xi\pi/2)} \\ &= {\pi \over \cosh\qty{\xi\pi\over 2}} \\ \\ &= \pi \operatorname{sech}\qty{\xi\pi\over 2 } .\end{align*}

• Let \(I\) be the integral over \({\mathbb{R}}\). Since \(f(x)\) is even, the original integral is \({1\over 2}I\).

• Write \(f(z) = e^{iz} / (z^2 + b^2)\). Take a semicircular contour \(\Gamma \coloneqq\gamma_1 + \gamma_2\) where \(\gamma_1\) is \([-R, R]\) on \({\mathbb{R}}\) and \(\gamma_2\) is the usual half-circle of radius \(R\).

• Claim: \(\int_{\gamma_2} f \overset{R\to\infty}\longrightarrow 0\), so \(\int_\Gamma \to \int_{\mathbb{R}}f(z)\).

  • Easy estimate, just be careful with the \(i\) in the exponent: \begin{align*} {\left\lvert {f} \right\rvert} = { {\left\lvert {e^{iz} } \right\rvert} \over {\left\lvert {z^2 + b^2} \right\rvert} } = {e^{-\Re z} \over {\left\lvert {z^2 + b^2} \right\rvert} } \leq {1\over {\left\lvert {z^2 + b^2} \right\rvert}} \overset{R\to\infty}\longrightarrow 0 .\end{align*}

• Compute \(\int_\Gamma f\) by residues: factor \(z^2 + b^2 = (z+ib)(z-ib)\), so the contour only contains the order 1 pole \(z_0 = ib\).

• Compute the residue: \begin{align*} \mathop{\mathrm{Res}}_{z=ib}f = \lim_{z\to ib} (z-ib) {e^{iz} \over (z+ib)(z-ib) } = { e^{iz} \over z+ib} \Big|_{z=ib} = {e^{i(ib)} \over 2ib} = {e^{-b} \over 2ib} .\end{align*}

• So the intermediate integral is \(I\) is \(2\pi i\) times this, i.e. \(I = \pi e^{-b} / b\).

• And the original integral is \({1\over 2}I = \pi e^{-b} \over 2b\).

Part 1: S&S claim this can be done without a calculation – maybe \({\left\lvert {\psi_a'(z)} \right\rvert}^2\) is constant on a circle of radius \(r\)…?

Part 2: \begin{align*} {1\over \pi} \iint_{\mathbb{D}}{\left\lvert {\psi_a'(z)} \right\rvert} \,dz &= {1\over \pi} \iint_{\mathbb{D}}{\left\lvert {1- {\left\lvert {a} \right\rvert}^2 \over (1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu z)^2 } \right\rvert} \,dz\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 \int_0^{2\pi} {1 \over {\left\lvert { 1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu re^{it} } \right\rvert}^2 } r\,dt\,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 \int_0^{2\pi} {1 \over (1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu re^{it} )(1 - a re^{-it}) } r\,dt\,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 \oint_{{{\partial}}{\mathbb{D}}} {1 \over (1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu rz )(1 - a r \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) } {r\over iz} \,dz\,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 \oint_{{{\partial}}{\mathbb{D}}} {1 \over (1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu rz )(1 - a r z^{-1}) } {r\over iz} \,dz\,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 {r\over i} \cdot \oint_{{{\partial}}{\mathbb{D}}} {1 \over (1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu rz )(z - a r) } \,dz\,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 {r\over i} \cdot 2\pi i \mathop{\mathrm{Res}}_{z=ar} {1 \over (1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu rz )(z - a r) } \,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 {r\over i} \cdot 2\pi i {1 \over (1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu rz ) }\Big|_{z=ar} \,dr\\ &= {2\pi \qty{ 1- {\left\lvert {a} \right\rvert}^2} \over \pi} \int_0^1 {r \over 1- {\left\lvert {a} \right\rvert}^2r^2 } \,dr\\ &= {1-{\left\lvert {a} \right\rvert}^2 \over {\left\lvert {a} \right\rvert}^2} \log\qty{1 \over 1 - {\left\lvert {a} \right\rvert}^2 } .\end{align*}

• Prove a more general statement: if \({\left\lvert {f(z)} \right\rvert} \leq M{\left\lvert {z} \right\rvert}^n\), then \(f\) is a polynomial of degree at most \(n\).

• Since \(f\) is entire, it is analytic everywhere, so \(f(z) = \sum_{k\geq 0}c_k z^k\) where \(c_k = f^{(k)}(0)/n!\) is given by the coefficient of its Taylor expansion about \(z=0\).

• Applying Cauchy’s estimate, on a circle of radius \(R\), \begin{align*} {\left\lvert {f^{(k)}(0)} \right\rvert} \leq { \sup_{\gamma}{\left\lvert {f(z)} \right\rvert} n! \over R^k} \leq {M{\left\lvert {z} \right\rvert}^n n! \over R^k} = {M R^n n! \over R^k} .\end{align*}

• So for \(k \geq n+1\), this goes to zero as \(R\to \infty\), so \({\left\lvert {f^{k}(0)} \right\rvert} = 0\) for all such \(k\).

• But then \(f\) is a power series annihilated by taking \(n+1\) derivatives, so it is a polynomial of degree at most \(n\).

\begin{align*} \int_\gamma z^n\,dz= \int_0^{2\pi} R^n e^{itn} \cdot iRe^{it} \,dt = R^{n+1} \int_0^{2\pi} e^{i(t+1)n}\,dt = { i R^{n+1} \over i(n+1) } \delta_{n+1 = 0} .\end{align*}

About a point \(a\) and \(R<{\left\lvert {a} \right\rvert}\), \begin{align*} \int_{{\left\lvert {z-a} \right\rvert} = R} z^n\,dz &= \int_0^{2\pi} (a + re^{it})^n \cdot ire^{it}\,dt\\ &= \int_0^{2\pi} \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} e^{it(n-k)} \cdot ire^{it}\,dt\\ &= i \int_0^{2\pi} \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} e^{it(n-k+1)} \,dt\\ &= i \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} \int_0^{2\pi} e^{it(n-k+1)} \,dt\\ &= i \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} \cdot 0 \\ &= 0 ,\end{align*} provided \(n\neq 0\), in which case \(\int_\gamma \,dz= 2\pi\).

For the third computation, this follows from partial fraction decomposition.

Note that \(G_z(\xi) \coloneqq{f(\xi) \over \xi - z}\) has a pole of order one at \(\xi = z\) and also a pole at \(\xi = \infty\). If \(z\in \Omega_1\), then \(\gamma\) encloses just the pole \(\xi = z\), so apply the residue theorem: \begin{align*} F(z) &\coloneqq{1\over 2\pi i}\oint_\gamma {f(\xi) \over \xi - z}\,d\xi\\ &= {1\over 2\pi i}\oint_\gamma G_z(\xi) \,d\xi\\ &= \mathop{\mathrm{Res}}_{\xi = z} G_z(\xi) \\ &= \lim_{\xi\to z} (\xi - z) G_z(\xi) \\ &= \lim_{\xi\to z} (\xi - z) {f(\xi) \over \xi-z} \\ &= \lim_{\xi\to z} f(\xi) \\ &= f(z) .\end{align*}

Now if \(z\in \Omega_2\), then \(\gamma\) encloses both \(\xi=z, \infty\), and is oriented negatively,so \begin{align*} F(z) &= {1\over 2\pi i} \oint_\gamma G_z(\xi) \,d\xi\\ &= -\qty{\mathop{\mathrm{Res}}_{\xi = z} G_z(\xi) + \mathop{\mathrm{Res}}_{\xi = \infty} G_z(\xi)}\\ &= -\qty{f(z) + \mathop{\mathrm{Res}}_{\xi = \infty} G_z(\xi)}\\ ,\end{align*} where the last line proceeds by the same calculation as above. It remains to compute the unknown residue. Residues at \(\xi = \infty\) are computed as residues at \(\xi =0\), and the change of variables \(G_z(\xi)\,d\xi\mapsto G_z(w) \,dw\) for \(w\coloneqq 1/\xi\) yields \(G_z(\xi)\,d\xi\to G_z\qty{1\over \xi}(-1/\xi^2)\,d\xi\). Thus \begin{align*} \mathop{\mathrm{Res}}_{\xi=\infty} G_z(\xi) &= -\mathop{\mathrm{Res}}_{\xi=0} G_z\qty{\xi^{-1}}\xi^{-2} \\ &= - \mathop{\mathrm{Res}}_{\xi=0} {f(\xi^{-1}) \over \xi^2(\xi^{-1}- z) } \\ &= - \mathop{\mathrm{Res}}_{\xi=0} {f(\xi^{-1}) \over \xi(1 - z\xi) } \\ &= -\lim_{\xi \to 0} {f(\xi^{-1}) \over 1-z\xi} \\ &= -\lim_{\xi \to 0}f(\xi ^{-1}) \\ &= -\lim_{\xi\to\infty} f(\xi) \\ &= -A .\end{align*} So combining this yields \begin{align*} F(z) = -\qty{f(z) - A} = -f(z) + A .\end{align*}

Green’s theorem: if \(\Omega\) is a domain with positively oriented boundary with \(u, v\) continuously differentiable in \(\mkern 1.5mu\overline{\mkern-1.5mu\Omega\mkern-1.5mu}\mkern 1.5mu\), then \begin{align*} \int_{{{\partial}}\Omega} u\,dx+ v\,dy= \iint_{\Omega}\qty{v_x - u_y}\,dx\,dy .\end{align*} Now use that if \(f = u+iv\) is analytic in a region, it satisfies Cauchy-Riemann: \begin{align*} u_x = v_y \qquad u_y = -v_x .\end{align*}

Now integrating \(f\): \begin{align*} \oint_{{{\partial}}\Omega} f(z) \,dz &= \oint_{{{\partial}}\Omega} (u+iv)(\,dx+ i\,dy)\\ &= \oint_{{{\partial}}\Omega} \qty{u\,dx- v\,dy} + i\oint_{{{\partial}}\Omega} \qty{v\,dx+ u\,dy} \\ &= \iint_\Omega\qty{v_x + u_y}\,dx\,dy+ \iint_\Omega\qty{u_x - v_y}\,dx\,dy\\ &= \iint_\Omega\qty{v_x -v_x }\,dx\,dy+ \iint_\Omega\qty{u_x - u_x}\,dx\,dy\\ &= 0 .\end{align*}

• By Cauchy’s integral formula, \(\int_{S^1} f = 2\pi i\)
• If \(p_j\) is any polynomial, then \(p_j\) is holomorphic in \({\mathbb{D}}\), so \(\int_{S^1} p_j = 0\).
• Contradiction: compact sets in \({\mathbb{C}}\) are bounded, so \begin{align*} {\left\lvert {\int f - \int p_j} \right\rvert} &\leq \int {\left\lvert {p_j - f} \right\rvert} \\ &\leq \int {\left\lVert {p_j - f} \right\rVert}_\infty \\ &= {\left\lVert {p_j - f} \right\rVert}_\infty \int_{S^1} 1 \,dz \\ &= {\left\lVert {p_j-f} \right\rVert}_\infty \cdot 2\pi \\ &\to 0 \end{align*} which forces \(\int f = \int p_j = 0\).

Let \(R> 10\), then by Cauchy: \begin{align*} 2\pi {\left\lvert {f'(z)} \right\rvert} &\leq \oint_{{\left\lvert {\xi} \right\rvert} = R} { {\left\lvert { f(\xi)} \right\rvert} \over {\left\lvert {\xi} \right\rvert}^2 } \,d\xi\\ &\leq \oint_{{\left\lvert {\xi} \right\rvert} = R} R^{-2} {\left\lvert {\xi} \right\rvert}^{1\over 2} \,d\xi\\ &= R^{-{3\over 2}} \cdot 2\pi R \\ &\sim R^{-{1\over 2}} \\ &\overset{R\to\infty}\longrightarrow 0 .\end{align*}

Toward applying Morera, let \(T \subseteq {\mathbb{C}}\setminus\gamma\) be a triangle, so that \(z\in T\) and \(w\in \gamma\) implies \(z-w\neq 0\). Then \begin{align*} \oint_T f(z) \,dz &= \oint_T \int_\gamma {g(w)\over w-z}\,dw\,dz\\ &= \int_\gamma \oint_T {g(w)\over w-z}\,dz\,dw\\ &= \int_\gamma g(w) \qty{ \oint_T {1 \over w-z}\,dz} \,dw\\ &= \int_\gamma g(w) \cdot 0 \,dw\\ &= 0 ,\end{align*} where the exchange of integrals is justified by compactness of \(\gamma, T\), and the inner integral vanishes because for a fixed \(w\in \gamma\), the function \(z\mapsto {1\over w-z}\) has a simple pole at \(w\), and so is holomorphic in \(\gamma^c\) and vanishes by Goursat.

Just reproducing the proof of holomorphicity in the Schwarz reflection theorem.

• Note \(f\) is continuous on \({\mathbb{C}}\) since analytic implies continuous (\(f\) equals its power series, where the partials sums uniformly converge to it, and uniform limit of continuous is continuous).
• Strategy: take \(D\) a disc centered at a point \(x\in {\mathbb{R}}\), show \(f\) is holomorphic in \(D\) by Morera’s theorem.
• Let \(\Delta \subset D\) be a triangle in \(D\).
• Case 1: If \(\Delta \cap{\mathbb{R}}= 0\), then \(f\) is holomorphic on \(\Delta\) and \(\int_\Delta f = 0\).
• Case 2: one side or vertex of \(\Delta\) intersects \({\mathbb{R}}\), and wlog the rest of \(\Delta\) is in \({\mathbb{H}}^+\).
  • Then let \(\Delta_{\varepsilon}\) be the perturbation \(\Delta + i{\varepsilon}= \left\{{z+ i{\varepsilon}{~\mathrel{\Big\vert}~}z\in \Delta}\right\}\); then \(\Delta_{\varepsilon}\cap{\mathbb{R}}= 0\) and \(\int_{\Delta_{\varepsilon}} f = 0\).
  • Now let \({\varepsilon}\to 0\) and conclude by continuity of \(f\) (???)
    • We want

      \begin{align*}     \int_{\Delta_{\varepsilon}} f = \int_a^b f(\gamma_{\varepsilon}(t)) \gamma_{\varepsilon}'(t)\,dt \overset{{\varepsilon}\to 0}\to \int_a^b f(\gamma(t)) \gamma_{\varepsilon}'(t)\,dt =\int_{\Delta}  f     \end{align*}

      where \(\gamma_{\varepsilon}, \gamma\) are curves parametrizing \(\Delta_{\varepsilon}, \Delta\) respectively.
    • Since \(\gamma, \gamma_{\varepsilon}\) are closed and bounded in \({\mathbb{C}}\), they are compact subsets. Thus it suffices to show that \(f(\gamma_{\varepsilon}(t)) \gamma_{\varepsilon}'(t)\) converges uniformly to \(f(\gamma(t))\gamma'(t)\).
    • ??
• Case 3: \(\Delta\) intersects both \({\mathbb{H}}^+\) and \({\mathbb{H}}^-\).
  • Break into smaller triangles, each of which falls into one of the previous two cases.

The main idea: \begin{align*} {\left\lvert {f'(z)} \right\rvert} &\leq {1\over 2\pi }\oint_R {{\left\lvert {f(\xi)} \right\rvert} \over {\left\lvert {\xi} \right\rvert}^2 } \,d\xi\\ &= {1\over 2\pi }\oint_R {{\left\lvert {f(\xi)} \right\rvert} } R^{-2} \,d\xi\\ &\leq {1\over 2\pi }\oint_R M R^{-2} \,d\xi\\ &= {1\over 2\pi} MR^{-2}\cdot 2\pi R \\ &= MR^{-1} \\ &\overset{R\to\infty}\longrightarrow 0 .\end{align*} So \(f'\equiv 0\).

Without loss of generality take \(a=0\). Since \(zf(z) \to A\) as \(z\to 0\), \(z=0\) is a simple pole of \(f\) and we can write \(f(z) = c_{-1}z^{-1}+ c_0 + c_1z + \cdots\). Then \begin{align*} \int_{\gamma_r} f(z)\,dz &= \int_{\gamma_r} \sum_{k\geq -1} c_k z^k \,dz\\ &= \sum_{k\geq -1} c_k \int_{\gamma_r} z^k \,dz\\ &= c_{-1}\int_{\gamma_r}{1\over z}\,dz\\ &= c_{-1}\int_{0}^{\beta_0} {1\over re^{i t}} ire^{it} \,dt\qquad z= re^{it}, \,dz= ire^{it} \,dt\\ &= i c_{-1}\int_{0}^{\beta_0} \,dt\\ &= i c_{-1}\beta_0 .\end{align*} Now use that \begin{align*} zf(z) = c_{-1} + c_0z + \cdots \overset{z\to 0}\longrightarrow c_{-1} = A ,\end{align*} so the integral is \(iA\beta_0\).

By the maximum modulus principle, \({\left\lvert {f} \right\rvert} \leq M\) in \(\mkern 1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu\). Since \(f\) has no zeros in \(\mkern 1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu\), \(g\coloneqq 1/f\) is holomorphic on \(D\) and continuous on \(\mkern 1.5mu\overline{\mkern-1.5muD\mkern-1.5mu}\mkern 1.5mu\). So the maximum modulus principle applies to \(g\), and \(M^{-1}\geq {\left\lvert {g} \right\rvert} = 1/{\left\lvert {f} \right\rvert}\), so \({\left\lvert {f} \right\rvert} \leq M\). Combining these, \({\left\lvert {f(z)} \right\rvert} = M\), so \(f(z) = \lambda M\) where \(\lambda\) is some constant with \({\left\lvert {\lambda} \right\rvert}=1\). This is on the unit circle, so \(\lambda = e^{i\theta}\) for some fixed angle \(\theta\).

Let \(M\coloneqq\sup_{z\in {{\partial}}U}{\left\lvert {f(z)} \right\rvert}\). If \(M=0\), then \(f\) must be the constant zero function, so assume \(M>0\).

Suppose toward a contradiction that there exists a \(z_0 \in U\) with \({\left\lvert {f(z_0)} \right\rvert} = M\). Note that the map \(z\mapsto {\left\lvert {z} \right\rvert}\) is an open in discs that don’t intersect \(z=0\). Since \(f\) is holomorphic, by the open mapping theorem \(f\) is an open map, so consider \(D_{\varepsilon}(z_0)\) a small disk not containing \(0\). Then \(f(D_{\varepsilon}(z_0))\) is open, and the composition \(z\mapsto f(z) \mapsto {\left\lvert {f(z)} \right\rvert}\) is an open map \(D_{\varepsilon}(z_0)\to {\mathbb{R}}\). Now if \(f\) is nonconstant, \({\left\lvert {f(D_{\varepsilon}(z_0))} \right\rvert} \supseteq(M-{\varepsilon}, M+{\varepsilon})\) contains some open interval about \(M\), which contradicts maximality of \(f\) at \(z_0\).

See notes for a proof using the mean value theorem.

Suppose toward a contradiction that \(f\) has no zeros in \(U\). Then \(g(z) \coloneqq 1/f(z)\) is holomorphic in \(U\). Now if \({\left\lvert {f(z)} \right\rvert} = C\) on \({{\partial}}U\), we have \({\left\lvert {g(z)} \right\rvert} = {\left\lvert {1\over f(z)} \right\rvert} = {1\over C}\) on \({{\partial}}U\), so \(\max_{z\in U} {\left\lvert {f(z)} \right\rvert} = C\) and \(\min_{{\left\lvert {f(z)} \right\rvert}} = {1\over C}\). Since \({\left\lvert {f(z)} \right\rvert}\) is constant on the boundary, we must have \(\max {\left\lvert {f(z)} \right\rvert} = \min {\left\lvert {f(z)} \right\rvert} = C\), so \(f\) is constant on \({{\partial}}U\). By the identity principle, \(f\) is constant on \(U\), a contradiction.

\begin{align*} {\left\lvert { f(z_0) } \right\rvert} &= {\left\lvert { {1\over 2\pi i} \oint_{{\left\lvert {z-z_0} \right\rvert} = R } {f(z) \over (z-z_0)^{n+1} } \,dz} \right\rvert} \\ &\leq {1\over 2\pi } \oint_{{\left\lvert {z-z_0} \right\rvert} = R } {\left\lvert {f(z)} \right\rvert} R^{-(n+1)} \,dz\\ &\leq {1\over 2\pi }\sup_{{\left\lvert {z-z_0} \right\rvert} = R} {\left\lvert {f(z)} \right\rvert} R^{-(n+1)} \cdot 2\pi R \\ &= \sup_{{\left\lvert {z-z_0} \right\rvert} = R} {\left\lvert {f(z)} \right\rvert} R^{-n} \\ &\leq (AR^k + B)R^{-n} \qquad \text{ if } z_0 = 0 \\ &= AR^{k-n} + BR^{-n} \\ &\to 0 ,\end{align*} provided \(k-n< 0\), so \(n>k\). Since \(f\) is entire, write \begin{align*} f(z) = \sum_{n\geq 0} f^{(n)}(0) {z^n\over n!} = \sum_{0\leq n\leq k} f^{(n)}(0) {z^n\over n!} ,\end{align*} making \(f\) a polynomial of degree at most \(k\).

Write \(S_\phi \coloneqq\left\{{0<\operatorname{Arg}(z) < \phi}\right\}\) and choose \(n\) large enough so that \begin{align*} {\mathbb{D}}\subseteq S \cup\zeta_n S \cup\zeta_n^2 S \cup\cdots\cup\zeta_{n}^{n-1}S ,\end{align*} i.e. so that the rotated sectors cover the disc. By uniform convergence of \(f\) to \(0\) on \(S\), choose \(r<1\) small enough so that \({\left\lvert {f(z)} \right\rvert} < {\varepsilon}\) for \({\left\lvert {z} \right\rvert} < r\) in \(S\). Note that \({\mathbb{D}}_r \subseteq \displaystyle\bigcup_{k=0}^{n-1} \zeta_n^k S_r\), where \(S_r \coloneqq\left\{{z\in S {~\mathrel{\Big\vert}~}{\left\lvert {z} \right\rvert} \leq r}\right\}\) is a subsector of radius \(r\).

By the MMP, let \(M\) be the maximum of \(f\) on \({\mathbb{D}}\), which is attained at some point on \(S^1\). Then \({\left\lvert {f} \right\rvert} < M\) on every \(\zeta_n^k S_r\). Now define \begin{align*} g(z) \coloneqq f(z) \prod_{k=1}^{n-1} f(\zeta_n^k z) \coloneqq f(z) \prod_{k=1}^{n-1}f_k(z) .\end{align*} Note that \({\left\lvert {f(z)} \right\rvert}\leq {\varepsilon}\) and \({\left\lvert {f_k(z)} \right\rvert} \leq M\), so \begin{align*} {\left\lvert {g(z)} \right\rvert}\leq {\varepsilon}\cdot M^{n-1} \overset{{\varepsilon}\to 0}\longrightarrow 0 .\end{align*} since \(M\) is a constant. So \(g(z) \equiv 0\) on \({\mathbb{D}}_r\), and by the identity principle, on \({\mathbb{D}}\). Thus some factor \(f_k(z)\) is identically zero. But if \(f(\zeta_n^k z)\equiv 0\) on \({\mathbb{D}}\), then \(f(z) \equiv 0\) on \({\mathbb{D}}\), since every \(z\in {\mathbb{D}}\) can be written as \(\zeta_n^k w\) for some \(w\in {\mathbb{D}}\).

Consider \begin{align*} f(z) \coloneqq\prod_{1\leq k \leq n} (w_k - z) .\end{align*} Then \(f\) is holomorphic and nonconstant on \({\mathbb{D}}\), so attains a maximum \(M\) on \(S^1\). Moreover, \({\left\lvert {f(z)} \right\rvert} = \prod {\left\lvert {w_k-z} \right\rvert}\) is exactly the product of distances from \(z\) to the \(w_k\). Moreover, since \({\left\lvert {f(0)} \right\rvert} = \prod{\left\lvert {w_k} \right\rvert} = 1\), we must have \(M>1\).

Now note that \(f(w_k) = 0\) and \(f\) is continuous in \({\mathbb{D}}\). So \({\left\lvert {f(z)} \right\rvert} \in [0, M] \subseteq {\mathbb{R}}\) where \(M>1\), so by the intermediate value theorem, \({\left\lvert {f(z)} \right\rvert} = 1\) for some \(z\).

Write \(f=u+iv\) where by assumption \(u\) is bounded. Both \(u\) and \(v\) are harmonic, so if \({\left\lvert {u} \right\rvert} \leq M\) on \({\mathbb{C}}\), then there is some disc where \({\left\lvert {u} \right\rvert} = M\) for some point in the interior. By the MMP for harmonic functions, \(u\) is constant on \({\mathbb{C}}\). So \(u_x, u_y = 0\), and by Cauchy-Riemann, \(v_x, v_y = 0\), so \(v'=0\) and \(v\) is constant, making \(f\) constant.

Consider \(g(z) \coloneqq e^{f(z)}\), then \({\left\lvert {g(z)} \right\rvert} = e^{\Re(z)}\) is entire and bounded and thus constant by Liouville’s theorem. So \(g'(z) = 0\), but on the other hand \(g'(z) = f'(z) e^{f(z)} = 0\), so \(f'(z) = 0\) and \(f\) must be constant since \(e^f\) is nonvanishing.

Computing the LHS: \begin{align*} \int_{[0, 2\pi]} {\left\lvert {f(re^{i\theta})} \right\rvert}^2 \,d\theta &= \int_{[0, 2\pi]} f(re^{i\theta}) \mkern 1.5mu\overline{\mkern-1.5muf(re^{i\theta}) \mkern-1.5mu}\mkern 1.5mu \,d\theta\\ &= \int_{[0, 2\pi]} \sum_{k\geq 0} c_k r^k e^{ik\theta} \sum_{j\geq 0} \mkern 1.5mu\overline{\mkern-1.5muc_j\mkern-1.5mu}\mkern 1.5mu r^j e^{-ij\theta} \,d\theta\\ &= \int_{[0, 2\pi]} \sum_{k,j\geq 0} c_k\mkern 1.5mu\overline{\mkern-1.5muc_j\mkern-1.5mu}\mkern 1.5mu r^{k+j} e^{i(k-j)\theta} \,d\theta\\ &= \sum_{k,j\geq 0} c_k\mkern 1.5mu\overline{\mkern-1.5muc_j\mkern-1.5mu}\mkern 1.5mu r^{k+j} \int_{[0, 2\pi]} e^{i(k-j)\theta} \,d\theta\\ &= \sum_{k,j\geq 0} c_k\mkern 1.5mu\overline{\mkern-1.5muc_j\mkern-1.5mu}\mkern 1.5mu r^{k+j} \chi_{i=j}\cdot 2\pi \\ &= \sum_{k\geq 0} c_k\mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu r^{2k} \cdot 2\pi \\ &= 2\pi \sum_{k\geq 0}{\left\lvert {c_k} \right\rvert}^2 r^{2k} ,\end{align*} where we’ve used that the series converges uniformly in its radius of convergence to commute sums and integrals.

Now supposing \({\left\lvert {f(z)} \right\rvert}\leq M\) for all \(z\in {\mathbb{C}}\), if \(f\) is entire then \(\sum_{k\geq 0} c_k z^k\) converges for all \(r\), so \begin{align*} \sum_{k\geq 0} {\left\lvert {c_k} \right\rvert}^2 r^{2k} = {1\over 2\pi }\int_{[0, 2\pi]} {\left\lvert {f(re^{i\theta})} \right\rvert}^2 \,d\theta\leq {1\over 2\pi}\int_{[0, 2\pi]} M^2 \,d\theta= M^2 .\end{align*} Thus for all \(r\), \begin{align*} {\left\lvert {c_0} \right\rvert}^2 + {\left\lvert {c_1} \right\rvert}^2 r^2 + {\left\lvert {c_2} \right\rvert}^2 r^{4} + \cdots \leq M^2 ,\end{align*} and taking \(r\to\infty\) forces \({\left\lvert {c_1} \right\rvert}^2 = {\left\lvert {c_2} \right\rvert}^2 = \cdots = 0\). So \(f(z) = c_0\) is constant.

• Strategy: By contradiction with Liouville’s Theorem
• Suppose \(p\) is non-constant and has no roots.
• Claim: \(1/p(z)\) is a bounded holomorphic function on \({\mathbb{C}}\).

  • Holomorphic: clear? Since \(p\) has no roots.

  • Bounded: for \(z\neq 0\), write

    \begin{align*}     \frac{P(z)}{z^{n}}=a_{n}+\left(\frac{a_{n-1}}{z}+\cdots+\frac{a_{0}}{z^{n}}\right)     .\end{align*}

  • The term in parentheses goes to 0 as \({\left\lvert {z} \right\rvert}\to \infty\)

  • Thus there exists an \(R>0\) such that

    \begin{align*}     {\left\lvert {z} \right\rvert} > R \implies {\left\lvert {P(z) \over z^n} \right\rvert} \geq c \coloneqq{{\left\lvert {a_n} \right\rvert} \over 2}     .\end{align*}

  • So \(p\) is bounded below when \({\left\lvert {z} \right\rvert} > R\)

  • Since \(p\) is continuous and has no roots in \({\left\lvert {z} \right\rvert} \leq R\), it is bounded below when \({\left\lvert {z} \right\rvert} \leq R\).

  • Thus \(p\) is bounded below on \({\mathbb{C}}\) and thus \(1/p\) is bounded above on \({\mathbb{C}}\).

• By Liouville’s theorem, \(1/p\) is constant and thus \(p\) is constant, a contradiction.

• Suppose \(f\) is entire and define \(g(z) \coloneqq{z \over f(z)}\).
• By the inequality, \({\left\lvert {g(z)} \right\rvert} \leq 1\), so \(g\) is bounded.
• \(g\) potentially has singularities at the zeros \(Z_f \coloneqq f^{-1}(0)\), but since \(f\) is entire, \(g\) is holomorphic on \({\mathbb{C}}\setminus Z_f\).
• Claim: \(Z_f = \left\{{0}\right\}\).
  • If \(f(z) = 0\), then \({\left\lvert {z} \right\rvert} \leq {\left\lvert {f(z)} \right\rvert} = 0\) which forces \(z=0\).
• We can now apply Riemann’s removable singularity theorem:
  • Check \(g\) is bounded on some open subset \(D\setminus\left\{{0}\right\}\), clear since it’s bounded everywhere
  • Check \(g\) is holomorphic on \(D\setminus\left\{{0}\right\}\), clear since the only singularity of \(g\) is \(z=0\).
• By Riemann’s removable singularity theorem, the singularity \(z = 0\) is removable and \(g\) has an extension to an entire function \(\tilde g\).
• By continuity, we have \({\left\lvert {\tilde g(z)} \right\rvert} \leq 1\) on all of \({\mathbb{C}}\)
  • If not, then \({\left\lvert {\tilde g(0)} \right\rvert} = 1+{\varepsilon}> 1\), but then there would be a domain \(\Omega \subseteq {\mathbb{C}}\setminus\left\{{0}\right\}\) such that \(1 < {\left\lvert {\tilde g(z)} \right\rvert} \leq 1 +{\varepsilon}\) on \(\Omega\), a contradiction.
• By Liouville, \(\tilde g\) is constant, so \(\tilde g(z) = c_0\) with \({\left\lvert {c_0} \right\rvert} \leq 1\)
• Thus \(f(z) = c_0^{-1}z \coloneqq cz\) where \({\left\lvert {c} \right\rvert}\geq 1\)

Thus all such functions are of the form \(f(z) = cz\) for some \(c\in {\mathbb{C}}\) with \({\left\lvert {c} \right\rvert}\geq 1\).

Since \(f\) is entire, take a Laurent expansion at \(z=0\), so \(f(z) = \sum_{k\geq 0} c_k z^k\) where \({2\pi i\over k!} c_k = f^{(k)}(0)\) by Cauchy’s integral formula. Take a Cauchy estimate on a disc of radius \(R>10\): \begin{align*} {\left\lvert {c_k} \right\rvert} &\leq {k!\over 2\pi}\int_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {f(\xi) \over (\xi - 0)^{k+1}} \right\rvert}\,d\xi\\ &\leq {k! \over 2\pi}\int_{{\left\lvert {z} \right\rvert} = R}{ {\left\lvert {\xi} \right\rvert}^{1\over 2} \over {\left\lvert {\xi} \right\rvert}^{k+1} }\,d\xi\\ &= {k! \over 2\pi} \cdot {1\over R^{k+{1\over 2}}}\cdot 2\pi R \\ &= { \mathsf{O}} (1/R^{k-{1\over 2}}) .\end{align*} So in particular, if \(k\geq 1\) then \(k-{1\over 2}>0\) and \(c_k = 0\). This forces \(f = c_0\) to be constant.

We have \({\left\lvert {f(z)} \right\rvert}\geq M\) for some \(M\), so \({\left\lvert {1/f(z)} \right\rvert} \leq M^{-1}\) is bounded, and we claim it is entire as well. This follows from the fact that \(1/f\) has singularities at the zeros of \(f\), but these are removable since \(1/f\) is bounded in every neighborhood of each such zero. So \(1/f\) extends to a holomorphic function. But now \(1/f =c\) is constant by Liouville, which forces \(f= 1/c\) to be constant.

Apply PFD and use that \(f\) is holomorphic to apply Cauchy’s formula over a curve of radius \(R\) enclosing \(a\) and \(b\): \begin{align*} \int_\gamma {f(z) \over (z-a)(z-b)}\,dz &= \int_\gamma f(z)\qty{{a-b \over z-a} + {b-a\over z-b} } \,dz\\ &= (a-b)^{-1}\int_\gamma {f(z) \over z-a} \,dz+ (b-a)^{-1}\int_\gamma {f(z) \over z-b}\,dz\\ &= (a-b)^{-1}\cdot 2\pi i f(a) + (b-a)\cdot 2\pi i f(b)\\ &= 2\pi i\qty{f(a) - f(b) \over a-b } .\end{align*} Since \(f\) is bounded, this number is finite and independent of \(R\), so taking \(R\to\infty\) preserves this equality. On the other hand, if \({\left\lvert {f(z)} \right\rvert}\leq M\), then we can estimate this integral directly as \begin{align*} I \leq \int_{{\left\lvert {z} \right\rvert} = R} {M \over {\left\lvert {R-a} \right\rvert} \cdot {\left\lvert {R-b} \right\rvert} } = {M\cdot 2\pi R \over {\left\lvert {R-a} \right\rvert} \cdot {\left\lvert {R-b} \right\rvert}} \ll {1\over R} \to 0 ,\end{align*} which forces \(f(a) =f(b)\). Since \(a, b\) were arbitrary, \(f\) must be constant.

Take a Laurent expansion at zero: \begin{align*} f(z) = \sum_{k\geq 0} c_k z^k,\qquad c_k = {1\over k!} f^{(k)}(0) = {1\over 2\pi i}\oint_{{\left\lvert {\xi} \right\rvert} = R} {f(\xi) \over \xi^{k+1}}\,d\xi .\end{align*} The usual estimate: \begin{align*} 2\pi i{\left\lvert {c_k} \right\rvert} \leq \oint_{{\left\lvert {\xi} \right\rvert} = R} R^{-(k+1)}{\left\lvert {f(\xi)} \right\rvert} \,d\xi &\leq \oint_{{\left\lvert {\xi} \right\rvert} = R}R^{-(k+1)} M R^2 \,d\xi\\ &= M R^{-(k-1)} \cdot 2\pi R \\ &= 2\pi M R^{-k+2} \\ &\overset{R\to\infty}\longrightarrow 0 ,\end{align*} provided \(-k+2<0 \iff k>2\).

\begin{align*} {\left\lvert {f^{(n)}(z)} \right\rvert} &= {\left\lvert { {1\over 2\pi i} \oint_\gamma {f(\xi) \over (\xi - z)^{n+1}} \,d\xi} \right\rvert} \\ &\leq {1\over 2\pi i} \oint_\gamma { {\left\lvert { f(\xi) } \right\rvert} \over {\left\lvert {\xi - z} \right\rvert}^{n+1}} \,d\xi\\ &\leq {1\over 2\pi i } \oint_\gamma {LR^m \over R^{n+1} } \,d\xi\\ &= {L\over 2\pi i} R^{m-(n+1)} \cdot 2\pi R \\ &= LR^{m-n} \\ &\overset{R\to\infty}\longrightarrow 0 \qquad \iff m-n<0 \iff n>m ,\end{align*} so \(f\) is a polynomial of degree at most \(m\).

Now if \(f\) is entire, \(g(z) \coloneqq e^{f(z)}\) is entire and \begin{align*} {\left\lvert {g(z)} \right\rvert} = {\left\lvert {e^{f(z)}} \right\rvert} = e^{\Re(f)} \leq e^M ,\end{align*} so \(g\) is an entire bounded function and thus constant by Liouville, making \(f\) constant. Why this is true: \begin{align*} e^{f} = C \implies e^f \cdor f' = 0 \implies f'\equiv 0 ,\end{align*} since \(e^f\) is nonvanishing, and \(f'\equiv 0\) implies \(f\) is constant.

Choose \({\left\lvert {z} \right\rvert}\) large enough so that \({\left\lvert {f(z)} \right\rvert}/{\left\lvert {z} \right\rvert}^n < {\varepsilon}\). Then write \(f(z) = \sum_{k\geq 0} c_k z^k\) and estimate \begin{align*} 2\pi {\left\lvert {c_k} \right\rvert} &\leq \oint_{{\left\lvert {z} \right\rvert} = R} {f(\xi) \over \xi^{k+1}}\,d\xi\\ &\leq \oint_{{\left\lvert {z} \right\rvert} = R} {{\varepsilon}{\left\lvert {\xi} \right\rvert}^n \over {\left\lvert {\xi} \right\rvert}^{k+1} } \,d\xi\\ &= {\varepsilon}R^{n-(k+1)} \cdot 2\pi R \\ &= {\varepsilon}C R^{n-k} \\ &\overset{{\varepsilon}\to 0}\longrightarrow 0 \end{align*} provided \(n-k \leq 0 \iff k\geq n\), since \({\varepsilon}\to 0\) forces \(R\to \infty\).

The easier version of this question: when \({\left\lvert {f} \right\rvert} \leq A{\left\lvert {z} \right\rvert}^N\), \(f\) is a polynomial of degree at most \(N\) by Cauchy’s integral formula: \begin{align*} {\left\lvert {f^{(n)}(z)} \right\rvert} &= {\left\lvert { {1\over 2\pi i} \oint_\gamma {f(\xi) \over (\xi - z)^{n+1}} \,d\xi} \right\rvert} \\ &\leq {1\over 2\pi i} \oint_\gamma { {\left\lvert { f(\xi) } \right\rvert} \over {\left\lvert {\xi - z} \right\rvert}^{n+1}} \,d\xi\\ &\leq {1\over 2\pi i } \oint_\gamma {AR^N \over R^{n+1} } \,d\xi\\ &= {A\over 2\pi i} R^{N-(n+1)} \cdot 2\pi R \\ &= AR^{N-n} \\ &\overset{R\to\infty}\longrightarrow 0 \qquad \iff N-n<0 \iff n>N .\end{align*}

Now rearrange the given equality \begin{align*} {\left\lvert {f(z) \over z^N} \right\rvert} \geq A \qquad {\left\lvert {z} \right\rvert} \implies {\left\lvert {z^N\over f(z)} \right\rvert} \leq A^{-1} .\end{align*} A priori, \(f\) is equal to its power series at \(z=0\), so \(f(z) = \sum_{k\geq 0} c_k z^k\). Since \({\mathbb{D}}_R\) is compact, \(f\) has finitely many zeros in this region, say \(\left\{{z_k}\right\}_{k\leq m}\). This set must be finite, since an infinite subset of a compact set has a limit point, and being zero on a set with a limit point implies being identically zero by the identity principle.

Define \begin{align*} p(z) \coloneqq\prod_{1\leq k\leq m} (z-z_k) = z^m + { \mathsf{O}} (z^{m-1}) ,\end{align*} the product of these roots. Increase \(R\) if necessary to ensure that \begin{align*} {\left\lvert {p(z)\over z^m} \right\rvert} < 1 \implies {\left\lvert {p(z)} \right\rvert} < {\left\lvert {z} \right\rvert}^m .\end{align*} Now define \begin{align*} G(z) \coloneqq{p(z) z^N \over f(z)} \implies {\left\lvert {G(z)} \right\rvert} = {\left\lvert {p(z) z^N\over f(z)} \right\rvert} = {\left\lvert {z^N\over f(z)} \right\rvert}\cdot {\left\lvert {p(z)} \right\rvert} \leq A^{-1}{\left\lvert {z} \right\rvert}^m .\end{align*}

Issue: this might not be entire? There could be poles at the zeros of \(f\) outside of \({\mathbb{D}}_R\)…

By the previous result, \(G\) is a polynomial of degree at most \(m\). Now consider leading terms: on one hand, \begin{align*} f(z) G(z) = p(z) z^N \sim \qty{z^m + \cdots }\cdot z^N = z^{N+m} + \cdots .\end{align*} On the other hand, \begin{align*} f(z) G(z) &= f(z) \qty{z^m + \cdots} \\ &\sim \sum_{k\geq 0} c_k z^{k+m} + z^{m-1}f(z) + \cdots \\ &= (z^m + \cdots + c_{N}z^{N+m} + \cdots) + z^{m-1}f(z) + \cdots ,\end{align*} and by the previous expression, this must be a polynomial of degree at most \(N+m\). This forces \(c_k = 0\) for all \(k> N\), otherwise these would contribute higher order terms.

Note: maybe not quite right!

Alternatively, note that the inequality can be rewritten as \begin{align*} {\left\lvert {G(z)} \right\rvert} \leq A^{-1}{\left\lvert {z} \right\rvert}^m \implies {\left\lvert {p(z)\over f(z)} \right\rvert} \leq A^{-1}{\left\lvert {z} \right\rvert}^{m-N} .\end{align*}

• If \(m-N = 0\), then \(p/f\) is an entire bounded function and thus constant, making \(p(z) = \lambda f(z)\) and \(f\) is a polynomial of degree exactly \(N\). -If \(m-N>0\), then \(p/f\) is a polynomial of degree at most \(m-N\) by the previous result. But \(p/f\) is a polynomial with no zeros, since \(Z_p = Z_f\), and the only nonvanishing polynomial is a constant, so again \(p = \lambda f\).
• If \(m-N<0\), then use the inequality \begin{align*} {\left\lvert {z^{N-m}p(z) \over f(z)} \right\rvert} \leq A^{-1} ,\end{align*} so the LHS is an entire bounded function and thus constant, so \(z^{N-m}p(z) = \lambda f(z)\). But the LHS is evidently a polynomial of degree \((N-m)+m = m\).

Note that the analogue of this problem where \({\left\lvert {f(z)} \right\rvert} \leq A {\left\lvert {z} \right\rvert}^N\) implies \(f\) is a polynomial of degree at most \(N\) is easy by the Cauchy estimate: \begin{align*} {\left\lvert {f(z)} \right\rvert} ={\left\lvert {\sum_{k\geq 0} c_k z^k } \right\rvert} \implies {\left\lvert {c_n} \right\rvert} = {\left\lvert {f^{(n)}(0)} \right\rvert} &= {\left\lvert {{n!\over 2\pi i }\int_\gamma {f(\xi) \over (\xi-a)^{n+1} } \,d\xi} \right\rvert} \quad \text{ at } a=0\\ &\leq {n!\over 2\pi }\int_\gamma {{\left\lvert {f(\xi)} \right\rvert} \over {\left\lvert {\xi} \right\rvert}^{n+1} } \,d\xi\\ &\leq {n!\over 2\pi }\int_\gamma {A {{\left\lvert {\xi} \right\rvert}^N } \over {\left\lvert {\xi} \right\rvert}^{n+1} } \,d\xi\\ &= {A n!\over 2\pi }\int_\gamma {{R ^N } \over R^{n+1} } \,d\xi\\ &= {An!\over 2\pi} \cdot {2\pi R \over R^{n+1-N}} \\ &= {An! \over R^{n-N}} \\ &\overset{R\to\infty}\longrightarrow 0 \quad \iff n-N>0 \quad\iff n>N ,\end{align*} so \(f(z) = \sum_{0\leq k\leq N} c_k z^k\).

For the case at hand, a solution I liked from MSE:

• Write \(g(z) \coloneqq f(1/z)\), so \(g\) has a singularity at \(z=0\). The claim is that this is a pole.

• It can’t be removable: \begin{align*} {\left\lvert {g(z)} \right\rvert} \geq A {\left\lvert {1\over z} \right\rvert}^n \to\infty \quad \text{ for } {\left\lvert {1/z} \right\rvert} \geq R \,\, (\iff {\left\lvert {z} \right\rvert} < 1/R) ,\end{align*} so \(g\) is unbounded near \(z=0\).

• It can’t be essential: if so, take the neighborhood of \(z=0\) given by \(U\coloneqq D_{1\over R}(0)\setminus\left\{{0}\right\}= \left\{{z{~\mathrel{\Big\vert}~}0< {\left\lvert {z} \right\rvert} < {1\over R} }\right\}\). Then \(g(U) \subseteq {\mathbb{C}}\) would be dense by Casorati-Weierstrass, but note that \(g(z) = w\in g(U) \implies {\left\lvert {w} \right\rvert} \coloneqq{\left\lvert {g(z)} \right\rvert} \geq A{\left\lvert {1/z} \right\rvert}^n\) since \({\left\lvert {z} \right\rvert}<1/R\), so \(g(U) \subseteq ({\mathbb{C}}\setminus D_{A\over R^n}(0))\) and in particular does not intersect the interior of \(D_{A\over R^n}(0)\).

• Since \(z=0\) is a pole, it has some finite order \(m\), so write \begin{align*} g(z) = \qty{c_{-m}z^{-m} + \cdots + c_{-1}z^{-1}} + \qty{c_0 + c_1 z + \cdots} \coloneqq p(1/z) + h(z) ,\end{align*} where \(p\) is polynomial of degree exactly \(m\) (since \(c_{-m} \neq 0\)) and \(h\) is entire. In particular, \(z=0\) is not a singularity of \(h\).

• Now \begin{align*} g(z) = p(1/z) + h(z) \implies f(z) = p(z) + h(1/z) .\end{align*}

• Then \begin{align*} f(z) - p(z) = h(1/z) \overset{{\left\lvert {z} \right\rvert}\to \infty}\longrightarrow c_0 \coloneqq h(0) ,\end{align*} since holomorphic functions are continuous.

• Then \(h\) is an entire function with a finite limit \(L\) at \(\infty\). \(h\) is bounded by \(c_0\) in a neighborhood \(U_\infty\) of \(\infty\) and takes on a maximum on \(U_\infty^c\) by compactness and the maximum modulus principle. So \(h\) is bounded on all of \({\mathbb{C}}\), and thus constant by Liouville, and thus \(h(1/z) = L\) for all \(z\).

• So \begin{align*} f(z) &= p(z) + h(1/z) = p(z) + c_0 \\ \implies f(z) &= (c_{-1}z + \cdots + c_{-m}z^m) + c_0 ,\end{align*} which is a polynomial of degree exactly \(m\coloneqq\deg p\).

• Why \(m \geq N\): if not, \(m<N\) so \(N-m > 0\). Then for large \(z\), \begin{align*} A \leq {\left\lvert {f(z) \over z^N} \right\rvert} &= {\left\lvert {c_0 + c_{-1}z + \cdots + c_{-m}z^m \over z^N} \right\rvert}\\ &= {\left\lvert { {c_0 \over z^N} + {c_{-1} \over z^{N-1}} + \cdots + {c_{-m} \over z^{N-m}} } \right\rvert} \\ &\overset{{\left\lvert {z} \right\rvert}\to\infty}\longrightarrow 0 ,\end{align*} since every term has a factor of \(z\) in the denominator. This contradicts \(A>0\). \(\contradiction\)

To clear up notation: write \(f(z) = u(x, y) + iv(x, y)\), here we’re assuming that \(u\) is polynomial in \(x\) and \(y\).

Note that \(f\) must have finitely many poles – either \(z=\infty\) is a pole or a removable singularity, and since poles are isolated, there is some \(R\gg 0\) such that all other poles of \(f\) are in \({\left\lvert {z} \right\rvert} \leq R\). The set \(P_f\) of poles is a closed set and \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}_R\mkern-1.5mu}\mkern 1.5mu\) is compact, so if \(P_f\) is infinite it has an accumulation point, contradicting that poles are isolated.

So enumerate \(P_f\) as \(\left\{{p_k}\right\}_{k\leq N}\), define \(g(z) \coloneqq\prod_{k\leq N}(z-p_k)\), and set \(F(z) \coloneqq g(z) f(z)\). Then \(F\) is an entire function, and the claim is that \(F\) is bounded and thus constant by Liouville. Proving the bound: take \({\left\lvert {z} \right\rvert} > R\), then \begin{align*} {\left\lvert {G(z)} \right\rvert} &= {\left\lvert {f(z)} \right\rvert} {\left\lvert {g(z)} \right\rvert} \\ &\leq M C {\left\lvert {z} \right\rvert}^N ,\end{align*} using that \(g\) is a polynomial of degree \(N\), so \({\left\lvert {g(z)\over z^N} \right\rvert}\to 1\) as \({\left\lvert {z} \right\rvert}\to \infty\) since \(g\) is monic. So after possibly increasing \(R\), we can choose \({\left\lvert {z} \right\rvert}\) large enough so that \({\left\lvert {g(z)\over z^N} \right\rvert} < C\) for, say, some constant \(C<2\). In any case, by a common qual question, if \({\left\lvert {G} \right\rvert} \in { \mathsf{O}} ({\left\lvert {z} \right\rvert}^N)\) for \({\left\lvert {z} \right\rvert}\) large enough then \(G\) is a polynomial of degree at most \(N\). Then \(f(z) \coloneqq G(z)/g(z)\) is a rational function.

Note that \(f\) has finitely many zeros: since \(f\) is unbounded, there is some \(R\) such that \(f({\mathbb{D}}_R^c) \subseteq {\mathbb{D}}^c\), so in particular \(f\) is nonvanishing on \({\mathbb{D}}_R^c\). So \(Z_f\) is a closed subset of a compact set, so is either finite or has an accumulation point. In the latter case, \(f\equiv 0\) by the identity principle, so suppose not.

Write \(Z_f = \left\{{z_k}\right\}_{k\leq n}\) for the \(n\) many zeros of \(f\), included with multiplicity, and set \begin{align*} \Phi(z) \coloneqq\prod_{k\leq n} (z-z_k), \qquad F(z) \coloneqq{\Phi(z) \over f(z) } .\end{align*} Now \(F\) is a nonvanishing entire function.

Given this, \(F\) is entire and bounded and thus constant by Liouville. So \(F(z) = c\), and as a result \(f(z) = c\Phi(z)\) which is a polynomial of degree \(n\).

Write \(Z_n \coloneqq\left\{{z\in {\mathbb{C}}{~\mathrel{\Big\vert}~}f^{(n)}(z) = 0 }\right\}\), then by hypothesis \(\displaystyle\bigcup_{n\geq 0} Z_n = {\mathbb{C}}\). A version of the Baire category theorem is that if \(X\) is a complete metric space and \(X\) is a countable union of closed sets, then at least one such set has a nonempty interior. Thus some \(Z_n\) has an interior point \(z_0\), and as a result there is some disc \({\mathbb{D}}_{\varepsilon}(z_0)\) on which \(f^{(n)}(z_0) \equiv 0\). This implies that \(f^{(k)}(z_0) \equiv 0\) on \({\mathbb{D}}_{\varepsilon}(z_0)\) for every \(k\geq n\), so \(f\) is a polynomial of degree at most \(n\).

Write \(f(z) = \sum_{k\leq n} c_k z^k\). Big: \(M(z) = c_nz^n\). Small: \(m(z) = f(z) - M(z) = \sum_{k\leq n-1} c_k z^k\).

Now use that \begin{align*} {\left\lvert {m(z) \over M(z)} \right\rvert} &\coloneqq{\left\lvert {c_n^{-1}\sum_{k\leq n-1} c_k z^{k-n}} \right\rvert} \\ &= {\left\lvert {c_n^{-1}\qty{ {c_1\over z^n} + {c_2\over z^{n-1} } + \cdots + {c_{n-1}\over z} }} \right\rvert} \\ &\overset{{\left\lvert {z} \right\rvert}\to\infty}\longrightarrow 0 ,\end{align*} so choose \(R\) large enough such that for \({\left\lvert {z} \right\rvert} \geq R\), \({\left\lvert {M(z)\over m(z)} \right\rvert} < 1\). Then on \({\left\lvert {z} \right\rvert} = R\), \begin{align*} {\left\lvert {m(z) \over M(z)} \right\rvert} < 1 \implies {\left\lvert {m(z)} \right\rvert} < {\left\lvert {M(z)} \right\rvert} \implies {\sharp}n = {\sharp}Z_{M} = {\sharp}Z_{M+m} = {\sharp}Z_{f} ,\end{align*} since \(c_n z^n\) has \(z=0\) as a root with multiplicity \(n\).

An estimate: write \(f(z) = \sum_{k\leq n} c_k z^k\) with \(c_n = 1\), then for \(R> 1\), on \({\left\lvert {z} \right\rvert} = R\) we have \begin{align*} {\left\lvert {f(z) - z^n} \right\rvert} &\leq \sum_{k\leq n-1} {\left\lvert { c_k z^k} \right\rvert} \\ &\leq \sum_{k\leq n-1} {\left\lvert { c_k} \right\rvert} R^k \\ &\leq \sum_{k\leq n-1} {\left\lvert { c_k} \right\rvert} R^{n-1} \\ &= R^{n-1} \sum_{k\leq n-1} {\left\lvert { c_k} \right\rvert} \\ &\coloneqq R^{n-1} C \\ &\leq R^n \\ &= {\left\lvert {z^n} \right\rvert} ,\end{align*} provided we can choose \(C<R\), but this is possible since \(\sum_{k\leq n-1}{\left\lvert {c_k} \right\rvert}\) is a constant. So \(n = {\sharp}Z_{z^n} = {\sharp}Z_f\).

On \({\left\lvert {z} \right\rvert} \leq 1\):

• Big: \(M(z) = -6z\)
• Small: \(m(z) = z^4 + 3\)

Then on \({\left\lvert {z} \right\rvert} = 1\), \begin{align*} {\left\lvert {m(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^4 + 3 = 4 < 6 = {\left\lvert {-6z} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so \(1 = Z_M = Z_f\) here.

On \({\left\lvert {z} \right\rvert} \leq 2\):

• Big: \(M(z) = z^4\)
• Small: \(m(z) = -6z+3\).

Then on \({\left\lvert {z} \right\rvert} = 2\), \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {-6z + 3} \right\rvert} \leq 6{\left\lvert {z} \right\rvert} + 3 = 15 < 16 = 2^4 = {\left\lvert {z} \right\rvert}^4 = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so \(4 = Z_M = Z_f\) here.

Thus there are \(4-1 = 3\) zeros in \(1 \leq {\left\lvert {z} \right\rvert} \leq 2\).

On \({\left\lvert {z} \right\rvert} \leq 1\):

• Big: \(M(z) = 12\)
• Small: \(m(z) = z^6 - 5z^3\)

For \({\left\lvert {z} \right\rvert} = 1\), \begin{align*} {\left\lvert {m(z)} \right\rvert} \coloneqq{\left\lvert {z^7-5z^3} \right\rvert}\leq {\left\lvert {z} \right\rvert}^7 + 5{\left\lvert {z} \right\rvert}^3 = 6 < 12 \coloneqq{\left\lvert {M(z)} \right\rvert} ,\end{align*} so \(0 = Z_M = Z_{f}\).

On \({\left\lvert {z} \right\rvert} \leq 2\),

• Big: \(M(z) = z^7\)
• Small: \(m(z) = -5z^3 + 12\)

On \({\left\lvert {z} \right\rvert} = 2\), \begin{align*} {\left\lvert {m(z)} \right\rvert} \coloneqq{\left\lvert {-5z^3 + 12} \right\rvert} \leq 5{\left\lvert {z} \right\rvert}^2 + 12 = 32 < 128 = 2^7 \coloneqq{\left\lvert {M(z)} \right\rvert} ,\end{align*} so \(7 = Z_M = Z_{f}\).

So \(f\) has 7 zeros in \(1 \leq {\left\lvert {z} \right\rvert} \leq 2\).

On \({\left\lvert {z} \right\rvert} < 1\):

• Small: \(m(z) = z^4+3\)
• Big: \(M(z) = -6z\)

On \({\left\lvert {z} \right\rvert} = 1\), \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {z^4+3} \right\rvert} \leq {\left\lvert {z} \right\rvert}^4 + 3 = 4 < 6 = {\left\lvert {-6z} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so \(Z_f = Z_M = 1\).

On \({\left\lvert {z} \right\rvert} < 2\):

• Small: \(-6z+3\)
• Big: \(z^4\)

On \({\left\lvert {z} \right\rvert} = 2\), \begin{align*} {\left\lvert {m(z)} \right\rvert} = \leq 6 + 3 = 9 < 2^4 = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so \(Z_f = Z_M = 4\).

Thus there are \(4-1=3\) zeros in \(1 \leq {\left\lvert {z} \right\rvert} \leq 2\).

Big: \(M(z) = -4z^3\). Small: \(m(z) = z^7 - 1\). Then on \({\left\lvert {z} \right\rvert} = 1\), \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {z^7-1} \right\rvert} \leq {\left\lvert {z} \right\rvert}^7 + 1 = 2 < 4 = {\left\lvert {-4z^4} \right\rvert} ,\end{align*} so \(f\) and \(M\) have the same number of zeros: three.

Big: \(M(z) = 4\) Small: \(m(z) = z^3 + 2z\). On \({\left\lvert {z} \right\rvert} = 1\), \begin{align*} {\left\lvert {m(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^3 + 2{\left\lvert {z} \right\rvert} = 1+2=3 < 4 = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so \(0 = Z_M = Z_{M+m} = Z_f\) in \({\mathbb{D}}\).

Big: \(M(z) = 3z^2\). Small: \(m(z) = z^3+bz + b^2\). Then on \({\left\lvert {z} \right\rvert} = 1\): \begin{align*} {\left\lvert {m(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^3 + b{\left\lvert {z} \right\rvert} + b^2 = 1 + b + b^2 < 3 = {\left\lvert {M(z)} \right\rvert} ,\end{align*} and \(M(z)\) has exact two roots in \({\mathbb{D}}\).

Key observation: \begin{align*} P_n(z) = \sum_{1\leq k\leq n-1} kz^{k-1} = {\frac{\partial }{\partial z}\,}Q_n(z) \qquad Q(z) \coloneqq\sum_{0\leq k \leq n} z^k .\end{align*} Note that \(Q(z) \to \sum_{k\geq 0} z^k = {1\over 1-z}\) uniformly on \({\left\lvert {z} \right\rvert} \leq R < 1\) since this power series has radius of convergence 1. Similarly \(P_n(z)\) converges uniformly to \({\frac{\partial }{\partial z}\,}{1\over 1-z} = {1\over (1-z)^2}\), so let \(P(z) \coloneqq{1\over (1-z)^2}\). Note that \(P\) is nonvanishing in \({\mathbb{D}}\).

Strategy:

• Small: \(m(z) = P(z) - P_n(z)\), look for an upper bound \({\left\lvert {m(z)} \right\rvert} < U\)
• Big: \(P(z)\), look for a lower bound \(L\) with \({\left\lvert {P(z)} \right\rvert} > L > U\).

Just by considering the geometry of circles of radius \(R < 1\) and \(1\) and measuring distances to the point \(1\), we can estimate \begin{align*} 0 < 1-R \leq {\left\lvert {1-z} \right\rvert} < 1+R < 2 \implies {\left\lvert {P(z)} \right\rvert} = {1\over {\left\lvert {1-z} \right\rvert}^2} \geq {1\over 2^2} = {1\over 4} .\end{align*}

Now fix \({\varepsilon}< {1\over 4}\) and use uniform convergence of \(P_n\to P\) to produce an \(N\) such that \(n\geq N\) implies \({\left\lVert {P-P_n} \right\rVert}_\infty < {\varepsilon}\) in \({\left\lvert {z} \right\rvert} \leq R\). Then on \({\left\lvert {z} \right\rvert} = R\), for \(n\geq N\), \begin{align*} {\left\lvert {m(z)} \right\rvert} \coloneqq{\left\lvert {P(z) - P_n(z)} \right\rvert} \leq {\left\lVert {P - P_n} \right\rVert}_\infty < {\varepsilon}< {1\over 4} \leq {\left\lvert {P(z)} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so \(0 = Z_P = Z_{P_n}\) by Rouché.

Write \(f(z) \coloneqq 6z^3 + 1 + e^z\).

• Small: \(m(z) = e^z + 1\)
• Large: \(M(z) = 6z^3\)
• The estimate: \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {e^z + 1} \right\rvert} \leq e^{\Re(z)} + 1 \leq e^{{\left\lvert {z} \right\rvert}} + 1 = e+1 < 6 = {\left\lvert {6z^3} \right\rvert} ,\end{align*} so \(3 = Z_M = Z_f\).

• Small: \(m(z) = z^6-3\)
• Big: \(M(z) = 4z^2 e^{z+1}\), which has two such zeros

Now estimate \(m\) from above: \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {z^6 - 3} \right\rvert} \leq {\left\lvert {z} \right\rvert}^6 + 3 = 4 .\end{align*} and \(M\) from below: \begin{align*} {\left\lvert {M(z)} \right\rvert} = {\left\lvert {4z^2 e^{z+1}} \right\rvert} = 4e{\left\lvert {z} \right\rvert}^4 e^{\Re(z)} = 4e e^{\Re(z)} \geq 4e e^{-1} = 4 ,\end{align*} which unfortunately isn’t quite enough. But equality occurs iff \(\Re(z) = -1\) on \(S^1\), so \(z=-1\), in which case \({\left\lvert {m(-1)} \right\rvert} = {\left\lvert {1-3} \right\rvert} = 2\), so the inequality is in fact strict. So \(2 = Z_M = Z_f\).

Take a contour \(\gamma_1 \coloneqq\left\{{it {~\mathrel{\Big\vert}~}t\in {\mathbb{R}}}\right\}\) and \(\gamma_2\coloneqq\left\{{Re^{it} {~\mathrel{\Big\vert}~}t\in [-\pi, \pi]}\right\}\).

• Big: \(M(z) = z^3 + 1\)
• Small: \(m(z) = -z\)

On \(\gamma_2\), we have \({\left\lvert {z} \right\rvert} = R\), so take \(R\) large enough that the following estimate holds: \begin{align*} {\left\lvert {M(z)} \right\rvert} = {\left\lvert {z^3 + 1} \right\rvert} \geq {\left\lvert { {\left\lvert {z} \right\rvert}^3 - 1} \right\rvert} = R^3 - 1 > R = {\left\lvert {m(z)} \right\rvert} = R .\end{align*} In particular, this works for \(R> 1\).

On \(\gamma_1\), note

• \({\left\lvert {M(z)} \right\rvert} = {\left\lvert { (it)^3 + 1 } \right\rvert} = {\left\lvert {1-it^3} \right\rvert}\)
• \({\left\lvert {m(z)} \right\rvert} = {\left\lvert {it} \right\rvert}\)

These can be interpreted geometrically: the former is the hypotenuse of a triangle and the latter is a leg, so \({\left\lvert {M(z)} \right\rvert} \geq {\left\lvert {m(z)} \right\rvert}\) will hold:

Now note that \(z^3 + 1\) has roots \(\omega_3, \omega_3^2, \omega_3^3=-1\) for \(\omega_k \coloneqq e^{i\pi\over k}\), and the first two are in the right half-plane. So \(2 = {\sharp}Z_M = {\sharp}Z_f\) by Rouché.

For the \(n=1\) case, \(f_1(z) = 0 \iff 1+z = 0 \iff z=-1\), so this has no roots in \({\mathbb{D}}\). For \(n=2\), factor \begin{align*} f_2(z) = 1 + z + z^2 = (z-\zeta_3^2)(z-\zeta_3^{-2}) ,\end{align*} using that \begin{align*} \zeta_3^2\cdot \zeta_3^{-2} = 1,\qquad -(\zeta_3^2 + \zeta_3^{-2}) = -2\Re(\zeta_3^2) = -2\cos\qty{2\pi \over 3} = 1 .\end{align*} Now use that \({\left\lvert {\zeta_3^{k}} \right\rvert} = 1\), which is not in \({\mathbb{D}}\).

For \(n\geq 3\): toward applying Rouche’s theorem, let \(M(z) = 1 + z\) and \(m(z) = {1\over 2}z^2 + \cdots + {1\over n!}z^n\). Note that on \({\left\lvert {z} \right\rvert} = 1\), \({\left\lvert {m(z)} \right\rvert} = 2\), and \begin{align*} {\left\lvert {m(z)} \right\rvert} &= {\left\lvert {\sum_{k\geq n+1} {z^k\over k!} } \right\rvert} \\ &\leq \sum_{k\geq n+1} { {\left\lvert {z} \right\rvert}^k \over k!} \\ &\leq \sum_{k\geq n+1} { 1 \over k!} \\ &= e^1 - \sum_{k\leq n} {1\over k!} .\end{align*} Suppose \(n\geq 3\), \begin{align*} {\left\lvert {m(z)} \right\rvert} < e - (1 + 1 + \cdots) \approx 0.718 < 2 ,\end{align*} then Rouché applies directly and \begin{align*} 0 = {\sharp}Z_M({\mathbb{D}}) ={\sharp}Z_{M+m}({\mathbb{D}}) \coloneqq{\sharp}Z_f({\mathbb{D}}) ,\end{align*} noting that \(M(z) = 0 \iff z= -1\), which isn’t contained in the open disc \({\mathbb{D}}\).

Note that \({\left\lvert {e^z} \right\rvert} = e^{\Re(z)}\), which is maximizes on \(S^1\) at \(z=1 \in {\mathbb{R}}\) and minimized at \(z=-1\). Write \(f(z) = e^z-az^n\), so solution correspond to zeros of \(f\).

Case 1: suppose \({\left\lvert {a} \right\rvert} > e\). Big: \(M(z) = az^n\). Small: \(m(z) = e^z\). On \({\left\lvert {z} \right\rvert} = 1\), \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {e^z} \right\rvert} = e^\Re(z) \leq e^1 < {\left\lvert {a} \right\rvert} = {\left\lvert {az^n} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so \(f\) has \({\sharp}Z_M = n\) zeros.

Case 2: suppose \({\left\lvert {a} \right\rvert} < 1/e\). Big: \(M(z) = e^z\). Small: \(m(z) = az^n\). On \({\left\lvert {z} \right\rvert} = 1\), \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {az^n} \right\rvert} = {\left\lvert {a} \right\rvert} < e^{-1}\leq e^{\Re(z)} = {\left\lvert {e^z} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*} and \(M\) has no zeros in \({\mathbb{D}}\) (and in fact none in \({\mathbb{C}}\)), so neither does \(f\).

More is true: this will hold for any disc of arbitrary radius \(R\), with \(n\) depending on \(R\). Fix \(R\), then use that \(P_n(z) \overset{n\to\infty}\longrightarrow e^z\) uniformly on the compact disc \({\left\lvert {z} \right\rvert} \leq R\). Consequently, setting \(g_n(z) \coloneqq{P_n(z)\over e^z}\), we have \(g_n(z) \to 1\) uniformly on this disc, for any \({\varepsilon}> 0\) this can be used to produce an \(n\gg 1\) such that \({\left\lvert { g_n(z) - 1 } \right\rvert} < {\varepsilon}\) for all \({\left\lvert {z} \right\rvert} \leq R\).

So take \({\varepsilon}\coloneqq 1\) and define \(h(z) \coloneqq 1\), then for \({\left\lvert {z} \right\rvert} = R\) \begin{align*} {\left\lvert {g_n(z) - 1} \right\rvert} < 1 = {\left\lvert {h(z)} \right\rvert} ,\end{align*} so by Rouché, \begin{align*} 0 = {\sharp}Z_{h} = {\sharp}Z_{h + (g_n - 1)} = {\sharp}Z_{g_n} ,\end{align*} since \(h\) has no zeros at all. Take \(R=10\) to get the stated result.

For \(P_n(z) - 1\), note that \(e^z-1=0\) has three solutions in \({\left\lvert {z} \right\rvert} < 10\), namely \(z=0, \pm 2\pi i\). We similarly have \(P_n(z)-1\to e^z-1\) uniformly, so on a disc of radius \(R\) choose \(n\) large enough so that \begin{align*} {\left\lvert {{P_n(z) -1 \over e^z - 1} - 1} \right\rvert} &< 1 \\ \implies {\left\lvert { (P_n(z) - 1) - (e^z-1) \over e^z-1} \right\rvert} &< 1 \\ \implies {\left\lvert { (P_n(z) - 1) - (e^z-1)} \right\rvert} &< {\left\lvert {e^z-1} \right\rvert} \\ \coloneqq{\left\lvert {m(z)} \right\rvert} &< {\left\lvert {M(z)} \right\rvert} ,\end{align*} so \begin{align*} 3 = {\sharp}Z_M = {\sharp}Z_{M+m} = {\sharp}Z_{P_n - 1} .\end{align*}

Write \(p(z) \coloneqq a_0 + \cdots + z^n\). Toward a contradiction, suppose not so that \({\left\lvert {p(z)} \right\rvert} < 1\) on \({\left\lvert {z} \right\rvert} = 1\). Then \begin{align*} {\left\lvert {f(z)} \right\rvert} < 1 = {\left\lvert {z} \right\rvert}^n \qquad \text{ on } {\left\lvert {z} \right\rvert} = 1 .\end{align*} Taking \(m(z) \coloneqq f(z)\) and \(M(z) \coloneqq-z^n\), we have \begin{align*} n = {\sharp}Z_M = {\sharp}Z_{M+m} = {\sharp}Z_{f(z) - z^n} \leq n-1 ,\end{align*} since \(f(z) - z^n\) is degree at most \(n-1\), a contradiction.

The boundary region is \(\left\{{1+it{~\mathrel{\Big\vert}~}t\in {\mathbb{R}}}\right\}\), write \(g(z) = ce^z-z\) so that fixed points of \(f\) are zeros of \(g\).

Big: \(M(z) = z\). Small: \(m(z) = ce^z\). Then for \(z=1+it\), \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {c} \right\rvert}e^{\Re(z)} < ce < 1 \leq \sqrt{1^2+t^2} = {\left\lvert {1+it} \right\rvert} = {\left\lvert {z} \right\rvert} ,\end{align*} so \(M\) and \(g\) have the same number of zeros, and \(M\) has a unique zero.

Consider \(f(z) \coloneqq z\sin(z) - a\).

Big: \(M(z) \coloneqq z\sin(z)\). Small: \(m(z) \coloneqq-a\). Use the following estimate: \begin{align*} {\left\lvert {z\sin(z)} \right\rvert}^2 &= {\left\lvert {z\over 2} \right\rvert}^2 {\left\lvert {e^{iz} - e^{-iz}} \right\rvert}^2 \\ &\geq {\left\lvert {z\over 2} \right\rvert}^2 {\left\lvert { {\left\lvert {e} \right\rvert}^{iz} } \right\rvert} - {\left\lvert { e^{-iz} } \right\rvert} }^2 \\ &= {\left\lvert {z\over 2} \right\rvert}^2 {\left\lvert {e^{-\Im(z)} - e^{\Im(z)} } \right\rvert} \\ &\overset{\Im(z)\to\infty}\longrightarrow \infty ,\end{align*} and so in particular a radius \(R\) can be chosen large enough so that \({\left\lvert {z\sin(z)} \right\rvert} > a\) for any \(a\). Thus for \({\left\lvert {z} \right\rvert} = R\), \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {a} \right\rvert} \leq {\left\lvert {z\sin(z)} \right\rvert} < {\left\lvert {M(z)} \right\rvert} \implies {\sharp}Z_{M} = {\sharp}Z_{M+m} = {\sharp}Z_f .\end{align*} To count the number of zeros of \(z\sin(z)\), note that this equals zero at \(z=0\) with multiplicity two and \(z= k\pi\) for \(k\in {\mathbb{Z}}\). Choosing \(R = {\pi \over 2} + n\pi\) for \(n\) large enough, there are exactly \(2n+2\) such zeros (with multiplicity) to \(z\sin(z)\), and thus \(2n+2\) zeros to \(z\sin(z) - a\). Now using that \(z\sin(z) - a\) has exactly \(2n+2\) real roots (??), this must be all of them.

Unsure how to find any roots of this thing, real or not!

The key step: getting the following inequality to work \begin{align*} {\left\lvert {az^n} \right\rvert} = {\left\lvert {a} \right\rvert}{\left\lvert {z} \right\rvert}^n < c \leq 1 = {\left\lvert { {\left\lvert {z} \right\rvert} - {\left\lvert {1} \right\rvert} } \right\rvert} \leq {\left\lvert {z+1} \right\rvert} .\end{align*} If this is true, then \(1 = {\sharp}Z_{z+1} = {\sharp}Z_f\). If \({\left\lvert {a} \right\rvert} < 2^n\), this holds because \({\left\lvert {a} \right\rvert}{\left\lvert {z} \right\rvert}^n < {1\over 2^n} 2^n = 1\), so taking \(c\coloneqq 1\) works.

Otherwise, suppose \({\left\lvert {a} \right\rvert} \geq 2^n\). Letting \(z_k\) be the roots of \(f\) and considering the constant coefficient, we have \begin{align*} a\prod_{k\leq n} z_k = 1 \implies {\left\lvert { \prod_{k\leq n} z_k } \right\rvert} = {\left\lvert {1\over a} \right\rvert} \leq 2^n ,\end{align*} so not every \(z_k\) can satisfy \({\left\lvert {z_k} \right\rvert} > 2\) and at least one is in \({\left\lvert {z} \right\rvert} \leq 2\).

Take a large semicircle

• \(\gamma_1 = [0, R]\)

• \(\gamma_2 = \left\{{Re^{it} {~\mathrel{\Big\vert}~}t\in [0, \pi/2]}\right\}\)

• \(\gamma_3 = i[0, R]\)

• \(\Delta\operatorname{Arg}(f, \gamma_1) = 0\): The only way \(f\circ \gamma_1\) can change argument is by changing sign, since it’s real valued. Use that \(f(0) = 10, f(1) = 11\) and \(f'(t) = 4t^3+6t-2 > 0\) so \(f\) is increasing on \([1, \infty)\). Then by Rouché, on \({\left\lvert {z} \right\rvert} = 1\) we have \({\left\lvert {z^4+2z^3-2} \right\rvert}\leq 5 < 10 = {\left\lvert {10} \right\rvert}\), so \(f\) has no zeros on \({\left\lvert {z} \right\rvert} \leq 1\).

• \(\Delta\operatorname{Arg}(f, \gamma_2) = 2\pi\): parameterize \(\gamma_2(t) = Re^{it}\), then \(f(\gamma(t)) \sim R^4e^{it}\) for large \(R\), which changes argument by \(2\pi\) for \(t\) in \([0, \pi/2]\).

• \(\Delta\operatorname{Arg}(f, \gamma_3) = 0\): check \(f(it) = t^4 + 10 + i(-2t^3-2t)\) and we let \(t\) range through \([0, R]\). For \(t>0\), the real part is strictly positive, so this can not wind about the origin.

By the argument principle, \({\sharp}Z_f = {1\over 2\pi} \Delta\operatorname{Arg}(f, \Gamma) = 1\).

It suffices to show there’s only one root in the open quadrant \(Q_1\), since they come in conjugate pairs. Assume that there are no roots on \({\mathbb{R}}\) or \(i{\mathbb{R}}\). Since polynomials are entire, the argument principle can be used to count zeros: \begin{align*} Z_f = {1\over 2\pi i}\int_\gamma { {\partial}^{\scriptsize \log} }f(z)\,dz= \Delta_\gamma \operatorname{Arg}(f) .\end{align*} To take the curve \(\gamma\) comprised of

• \(\gamma_1 = [0, R]\),
• \(\gamma_2 = Re^{it}\) for \(t\in [0, \pi/2]\)
• \(\gamma_3 = i[0, R]\).

Then

• \(\Delta_{\gamma_1}\operatorname{Arg}(f) = 0\), since \(f({\mathbb{R}}_{\geq 0}) \subseteq {\mathbb{R}}_{\geq 0}\).
• \(\Delta_{\gamma_2}\operatorname{Arg}(f) = 4\cdot {\pi\over 2} = 2\pi\) since \(f\sim z^4\) for large \(R\).
• \(\Delta_{\gamma_3}\operatorname{Arg}(f)\): consider \begin{align*} f(it) = t^4 - it^3 -2it + 10 = t^4\qty{1 - it^{-1}-2it^{-2} +10t^{-4}} \\ \implies \operatorname{Arg}(f(it)) \sim \operatorname{Arg}(t^4) =0 .\end{align*}

So \(\Delta_\gamma \operatorname{Arg}(f) = 1\), meaning there is one zero enclosed by \(\gamma\) for \(R\) large enough. As \(R\to \infty\), this covers \(Q_1\).

Write \(g(z) \coloneqq f(z) - w_0\), then \(g\) is holomorphic on \(D\) and thus \(w_0\) is an isolated zero. Choose \(\delta\) small enough so that \(g\) is nonvanishing on \({\mathbb{D}}_\delta(z_0)\setminus\left\{{ z_0 }\right\}\). Let \begin{align*} \gamma \coloneqq\left\{{{\left\lvert {\xi - z_0} \right\rvert} = \delta }\right\}= {{\partial}}{\mathbb{D}}_{\delta}(z_0) .\end{align*} Choose \({\varepsilon}< \inf\left\{{w\in f(\delta)}\right\}\) so that \({\left\lvert {f(z) - w_0} \right\rvert} > {\varepsilon}\) in \({\mathbb{D}}_{\varepsilon}(w_0)\setminus\left\{{ w_0 }\right\}\) for every \(z\in \gamma\). Let \begin{align*} \gamma' \coloneqq{{\partial}}{\mathbb{D}}_{{\varepsilon}}(w_0) = \left\{{{\left\lvert {z-w_0} \right\rvert} = {\varepsilon}}\right\} ,\end{align*} and define the solution counting function: \begin{align*} F(w) \coloneqq{1\over 2\pi i} \oint_{\gamma'} { {\partial}^{\scriptsize \log} }(g(z)) \,dz = {1\over 2\pi i } \oint_{\gamma'} {g'(z)\over g(z) }\,dz = {1\over 2\pi i} \oint_{\gamma'} {f'(z)\over f(z) - w} \,dz ,\end{align*} which counts the zeros of \(g\) (since it has no poles) and consequently the number of solutions to \(f(z) = w\) in \({\mathbb{D}}_{\varepsilon}(w_0)\). This is now a continuous integer valued function on \({\mathbb{D}}_{\varepsilon}(w_0)\), and is thus constant. Since \(f(z_0) = w_0\) with \(z_0\) enclosed by \(\gamma\) and \(w_0\) enclosed by \(\gamma'\), the constant is exactly the multiplicity of the zero of \(f(z) - w_0\) at \(z_0\), which is \(m\).

Note that \(f\) is holomorphic on \({\mathbb{D}}\) and \(S^1\), since the poles are at \(1/\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5mu\) and if \({\left\lvert {a_l} \right\rvert} < 1\) then \({\left\lvert {\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5mu} \right\rvert} > 1\). Fix \(b\), then define \(g_w(z) \coloneqq f(z) - w\) and form the solution counting function \begin{align*} F(w) \coloneqq{1\over 2\pi i}\oint_{S^1} { {\partial}^{\scriptsize \log} }g_w(z) \,dz = {1\over 2\pi i} \oint_{S^1} {f'(z) \over f(z)-w}\,dz .\end{align*} Start by computing \(F(0)\). \begin{align*} F(0) &= {1\over 2\pi i }\oint_{S^1} { {\partial}^{\scriptsize \log} }\prod_{1\leq k\leq n} \psi_{a_k}(z) \,dz\\ &= {1\over 2\pi i }\oint_{S^1} \sum_{1\leq k\leq n} { {\partial}^{\scriptsize \log} }\psi_{a_k}(z) \,dz\\ &= {1\over 2\pi i }\oint_{S^1} \sum_{1\leq k\leq n} \qty{1-{\left\lvert {a_k} \right\rvert}^2 \over (1-\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5mu z)^2} \qty{z-a_k \over 1-\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5mu z}^{-1}\,dz\\ &= {1\over 2\pi i }\oint_{S^1} \sum_{1\leq k\leq n} {1-{\left\lvert {a_k} \right\rvert}^2 \over (z-a_k)( 1-\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5muz) } \,dz\\ &= {1\over 2\pi i } \sum_{1\leq k\leq n} \oint_{S^1} {1-{\left\lvert {a_k} \right\rvert}^2 \over (z-a_k)( 1-\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5muz) } \,dz\\ &= {1\over 2\pi i } \sum_{1\leq k\leq n} 2\pi i \\ &= n ,\end{align*} where we’ve used that the integrand has a simple pole at \(a_k\) since \(1/\mkern 1.5mu\overline{\mkern-1.5mua_k\mkern-1.5mu}\mkern 1.5mu\in {\mathbb{D}}^c\). So the equation \(f(z) = 0\) has \(n\) solutions. Now use that \(F\) is a continuous function of \(w\) on \({\mathbb{D}}\) and integer valued, thus constant. So \(F(w) = n\) for any \(w\), meaning \(f(z) = w\) has \(n\) solutions in \({\mathbb{D}}\) for every \(w\).

Alternative: \(F\) continuously depends on the \(a_k\), so send them all to zero to get \(f(z) = z^n\) which trivially has \(n\) zeros.

By Morera’s theorem, it suffices to show \(\int_T f = 0\) for all triangles \(T \subseteq {\mathbb{D}}\). Noting that \(T\) is closed and bounded and thus compact, \(f_n\to g\) uniformly on \(T\). Since the \(f_n\) are holomorphic, \(\int_T f_n = 0\) for all \(n\), and thus \begin{align*} \int_T g = \int_T \lim f_n = \lim_n \int_T f_n = \lim_n 0 = 0 ,\end{align*} where moving the limit through the integral is justified by uniform convergence.

By Fubini: \begin{align*} \oint_T g(z)\,dz &\coloneqq\oint_T \int_{\mathbb{R}}f(t)e^{-izt} \,dt\,dz\\ &\coloneqq\int_{\mathbb{R}}\oint_T f(t)e^{-izt} \,dz\,dt\\ &\coloneqq\int_{\mathbb{R}}f(t) \qty{ \oint_T e^{-izt} \,dz} \,dt\\ &\coloneqq\int_{\mathbb{R}}f(t) \cdot 0 \,dt\\ &= 0 ,\end{align*} where the inner integral vanishes because \(z\mapsto e^{-izt}\) is entire by Goursat’s theorem. So \(g\) is entire by Morera.

Write \({\mathbb{D}}_{\varepsilon}(z_0)\) for a disc in which \(f\) is bounded, say by \({\left\lvert {f} \right\rvert}\leq M\) here. Note that if \(z_0\) is not in the region enclosed by \(\Delta\), then \(\int_\Delta f = 0\) since \(f\) is holomorphic throughout \(\Delta\), so suppose \(z_0\) is in this region.

Since \(f\) is holomorphic in \(\Omega\setminus\left\{{ z_0 }\right\}\), \(\Delta\) can be deformed to \({{\partial}}{\mathbb{D}}_{\varepsilon}(z_0)\) without changing the integral. So \begin{align*} {\left\lvert { \oint_\Delta f } \right\rvert} &= {\left\lvert {\oint_{{{\partial}}{\mathbb{D}}_{\varepsilon}(z_0)} f } \right\rvert} \\ &\leq \oint_{{{\partial}}{\mathbb{D}}_{\varepsilon}(z_0)} {\left\lvert {f} \right\rvert} \\ &\leq \oint_{{{\partial}}{\mathbb{D}}_{\varepsilon}(z_0)} M \\ &= M\cdot 2\pi {\varepsilon}\\ &\overset{{\varepsilon}\to 0}\longrightarrow 0 ,\end{align*} noting that the bound \(M\) is constant and remains an upper bound on smaller discs by the maximum modulus principle.

By Morera’s theorem, it suffices to show \(\int_\Delta f(z) \,dz= 0\) for all triangles \(\Delta \subseteq \gamma^c\). Claim: \begin{align*} \int_\Delta f(z) \,dz &= \int_\Delta \int_\gamma {F(w) \over w-z} \,dw\,dz\\ &= \int_\gamma \int_\Delta {F(w) \over w-z} \,dz\,dw\\ &= \int_\gamma F(w) \qty{ \int_\Delta {1 \over w-z} \,dz} \,dw\\ &= \int_\gamma F(w) \cdot 0 \,dw\\ &= 0 .\end{align*}

That the inner integral is zero follows from the fact that the function \(z\mapsto {1\over w-z}\) is holomorphic on \(\gamma^c\), since it has only a simple pole at \(w\) where \(w\in \gamma\) is fixed.

That the interchange of integrals is justified follows from Fubini’s theorem: these are continuous functions on compact sets, which are uniformly bounded and thus Lebesgue measurable and integrable.

The claim is that \(f\) is complex differentiable, thus smooth, thus holomorphic and equal to its Taylor series expansion. The quick justification: \begin{align*} {\frac{\partial }{\partial z}\,} f(z) &= {\frac{\partial }{\partial z}\,} \int_\gamma {F(w) \over w-z}\,dw\\ &= \int_\gamma {\frac{\partial }{\partial z}\,} {F(w) \over w-z} \,dw\\ &= \int_\gamma {F(w) \over (w-z)^2} \,dw ,\end{align*} where differentiating through the integral is justified since the integrand is a continuous function of \(z\) on \(\gamma\) since \(w\neq z\) on \(\gamma\), and \(\gamma\) is a compact set.

Slightly more rigorously, one can equivalently pass a limit through the integral to show that the defining limit exists: \begin{align*} f(z+h) - f(z) &= \int_\gamma {F(w) \over w+h-z} \,dw- \int_\gamma {F(w) \over w-z}\,dw\\ &= \int_\gamma {(w-z)F(w) - (w+h-z)F(w) \over (w+h-z)(w-z) } \,dw\\ &= \int_\gamma F(w) {h \over (w+h-z)(w-z)} \,dw\\ &\overset{h\to 0}\longrightarrow \int_\gamma {F(w) \over (w-z)^2}\,dw ,\end{align*} since the term involving \(h\) goes to 1.

Without loss of generality assume \(w_0 = 0\). If \({\left\lvert {f(z)} \right\rvert} \leq M\) for \({\left\lvert {z} \right\rvert} < {\varepsilon}\), pick \(T\) contained in \({\mathbb{D}}_{\varepsilon}\), then \begin{align*} {\left\lvert {\oint_T f(z)\,dz} \right\rvert} \leq \oint_T {\left\lvert {f(z)} \right\rvert}\,dz\leq \oint_T M \,dz= M \mu(T) ,\end{align*} and by homotopy invariance, this integral doesn’t change as \(T\) is perturbed. Shrinking \(T\), we can make \(\mu(T)\to 0\).

Use that every element of \(\mathop{\mathrm{Aut}}({\mathbb{D}})\) is of the form \(f(z) = \lambda\psi_a(z)\), and the region \(G\) is conformally equivalent to \({\mathbb{D}}\). Thus every element of \(\mathop{\mathrm{Aut}}(G)\) can be conjugated to an element of \(\mathop{\mathrm{Aut}}({\mathbb{D}})\) by a conformal map \(F: G\to {\mathbb{D}}\). One such map is gotten by rotating \(z\mapsto iz\) to get \({\mathbb{H}}\cap{\left\lvert {z} \right\rvert}>1\), then applying a Joukowski map \(z\mapsto z+z^{-1}\) to get \({\mathbb{H}}\). So \begin{align*} f\in \mathop{\mathrm{Aut}}(G) \implies f = F^{-1}\circ \psi_{a} \circ F \text{ for some } \psi_a \in \mathop{\mathrm{Aut}}({\mathbb{D}}) .\end{align*}

Part 1: this is a bigon with vertices \(0, \infty\), so send \(0\to\infty\) with \(1/z\). Orient \(i{\mathbb{R}}\) and the circle \(S\) positively, note that both will be mapped to generalized circles. To find the resulting region, use handedness – it’s on the right of \(i{\mathbb{R}}\) and the right of \(S\). The map preserves \(i{\mathbb{R}}\) and as \(t\) traces out \((-\infty, 0^-, 0^+, \infty)\), \(f(it)\) traces out \((0^+, \infty, -\infty, 0^-)\), so this preserves the orientation of \(i{\mathbb{R}}\). For \(S\), let \(z_0\coloneqq{1\over 2}(1+i)\in S\), then \(f(z_0) = 1-i\). So the arc \((1,z_0, 0)\) maps to \((1, 1-i,\infty)\), so this is a vertical line through \(\Re(z) = 1\) oriented downward. The region is to the right of \(S\), so we have

The rest is standard:

• Dilate and rotate to \(0<\Im(z) < \pi\) using \(z\mapsto i\pi z\).
• Exponentiate using \(z\mapsto e^z\)

  \to

  get \({\mathbb{H}}\).
• Apply the Cayley map \(z\mapsto {z-i\over z+i}\) to get \({\mathbb{D}}\).

Part 2: a bigon with vertex \(1\), i.e. a lune. Send \(1\to \infty\) with \(f(z) \coloneqq{1\over z-1}\), and check that

• \(1\mapsto \infty\)
• \({1\over 2}(1+i) \mapsto -(1+i)\)
• \(0\mapsto -1\)
• \(i\mapsto -{1\over 2}(1+i)\)
• \(-1\mapsto -{1\over 2}\)

By tracking tangent/normal vectors, this results in the region \(-1<\Re(z) < -{1\over 2}\):

The rest is standard:

• Translate to the right by \(z\mapsto z+{1\over 2}\) to get \(-{1\over 2}<\Re(z) < 0\).
• Rotate and dilate by \(z\mapsto -2i\pi z\) to get \(0<\Im(z) < \pi\)
• Exponentiate by \(z\mapsto e^z\) to get \({\mathbb{H}}\),
• Cayley map \(z\mapsto {z-i\over z+i}\) to get \({\mathbb{D}}\).

Part 3: a bigon in \({\mathbb{H}}\) with vertices \(\pm 1\), with an arc passing through \(z_3 \coloneqq i(\sqrt{2} - 1)\). Take \(z\mapsto {z+1\over z-1}\) to obtain

• \(-1\mapsto 0\)
• \(1\mapsto \infty\)
• \(0\mapsto -1\)

Orienting the bigon positively, we have \((-1, 0, 1)\mapsto (0, -1, \infty)\), i.e. the real axis oriented from \(+\infty\to-\infty\). Similarly \((1, z_3, -1)\mapsto (\infty, \omega_4^3, 0)\), which is a line passing through \(\omega_4^3\), oriented from \(Q_3\to Q_1\). Since the original region was on the left of both curves, we get

Now

• Flip this to \(Q_1\) with \(z\mapsto -z\) to get \(0<\operatorname{Arg}(z) < \pi/4\).
• Rotate clockwise with \(z\mapsto e^{-i\pi\over 8}\) to get \(-\pi/8<\operatorname{Arg}(z) < \pi/8\).
• Dilate the argument to a half-plane with \(z\mapsto z^{\pi\over 2\theta_0}\) where \(\theta_0 = \pi/8\) to get \(-\pi/2<\operatorname{Arg}(z) < \pi/2\).
• Rotate with \(z\mapsto iz\) to get \({\mathbb{H}}\).
• Cayley map, \(z\mapsto {z-i\over z+i}\).

Part 4: See part 5. The critical step is a Blaschke map \(\psi_a\) which sends \(a\to 0\). For \(a\in {\mathbb{R}}\), \(\psi_a({\mathbb{R}}) = {\mathbb{R}}\) and this will map the partial slit from \(a\) to the boundary to a usual slit from \(0\) to the boundary.

Part 5: Dealing with the slit:

• Use a Blaschke factor to send \(a\coloneqq-1/2\to 0\), so \(z\mapsto {a-z\over 1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu z}\). Checking that \((-1/2, 0, 1)\to (0, -1/2, -1)\), the image is \({\mathbb{D}}\setminus(-1, 0]\).
• Rotate with \(z\mapsto e^{-i\pi}z\) to get \({\mathbb{D}}\setminus[0, 1)\).
• Unfold with \(z\mapsto z^{1\over 2}\) to get \({\mathbb{D}}\cap{\mathbb{H}}\), noting that the slit becomes \([-1, 1]\) and is erased here.
• Use \(z\mapsto -1/z\) to get \({\mathbb{D}}^c \cap{\mathbb{H}}\).
• Use the Joukowski map \(z\mapsto z+z^{-1}\) to map to \(Q_{34}\)
• Use \(z\mapsto -z\) to get \({\mathbb{H}}\).

It’s well known that \(z\mapsto \sin(z)\) sends \(-\pi/2<\Re(z) < \pi/2\) with \(\Im(z) > 0\) to \({\mathbb{H}}\):

So take \(z\mapsto \arcsin(z)\).

This is a lune-type region:

The usual strategy is to blow up the tangency, so send \(1\to\infty\) with \begin{align*} f(z) \coloneqq{1\over z-1} .\end{align*}

From here, it’s a standard exercise. In steps:

• Map \(R\) to the vertical strip \(-1< \Re(z) < -{1\over 2}\) using \(z\mapsto {1\over z-1}\).
• Shift using \(z\mapsto z+{1\over 2}\) to send this to \(-{1\over 2}< \Re(z) < 0\).
• Rotate using \(z\mapsto -iz\) to get \(0<\Im(z) < {i\over 2}\), a horizontal strip.
• Dilate using \(z\mapsto 2\pi z\), which sends \({i\over 2}\to \pi i\), so the resulting region is \(0 < \Im(z) < \pi\).
• Apply \(z\mapsto e^z\) to map the horizontal strip to \({\mathbb{H}}\).
• Apply the Cayley map \(z\mapsto {z-i\over z+i}\) to map \({\mathbb{H}}\to{\mathbb{D}}\).

The region:

Note that you can find that \(i, -i\) are the intersection points by noting that \(i{\mathbb{R}}\) is the perpendicular bisector through the line segment connecting the centers of the circles, then expanding \({\left\lvert {z-1} \right\rvert}^2 = (x-1)^2 + y^2 = 2\) and setting \(x=0\) to get \(y=\pm i\).

First rotate this by \(\pi/2\):

Call the upper circle \(C_1\) and the lower \(C_2\). Send \(-1\to 0, 1\to \infty\) by taking \begin{align*} f(z) \coloneqq{z+1\over z-1} .\end{align*} This sends the circles to lines through zero, and the lune now spans a triangular sector. Finding the angles of the lines: write \(c\coloneqq 1 + \sqrt{2}\). Note that \(f\) fixes \({\mathbb{R}}\), so the image regions are symmetric about \({\mathbb{R}}\), and it suffices to find the angle of the line \(f(C_1)\). Note that \(C_1 \cap i{\mathbb{R}}= ic\), so we compute \(\operatorname{Arg}(f(ic))\): \begin{align*} f(ic) &= {ic+1\over ic-1} \\ &= {(1+ic)^2 \over c^2 + 1} \\ &= {-1 + c^2 - 2ic \over c^2 + 1} \\ &= {c^2-1 \over c^2+1} -i {2c\over c^2+1} ,\end{align*} so \begin{align*} \operatorname{Arg}(f(ic)) &= \arctan\qty{ 2c\over -1 + c^2} \\ &= \arctan\qty{2(1+\sqrt 2) \over 1 - (3 + \sqrt 2)} \\ &= \arctan(-1) \\ &= {\pi\over 4} \text{ or } {3\pi \over 4} .\end{align*}

Thus \(f(C_1) = \left\{{te^{-i\pi\over 4} {~\mathrel{\Big\vert}~}t\in (-\infty, \infty)}\right\}\). Note that \(f(ic)\in Q_4\), since \(c^2-1, c^2+1 > 0\) and \({-2c\over c^2+1} < 0\). For the orientation of \(f(C_1)\), note that \((1, ic, -1) \mapsto (\infty, f(ic), 0)\), so the line is oriented from \(Q_4\) to \(Q_2\).

A similar computation shows \begin{align*} f(-ic) = {c^2-1\over c^2+1} + i{2\over c^2+1} \in Q_1 ,\end{align*} and \((-1, -ic, 1)\mapsto (0, f(-ic), \infty)\), so \(f(C_2)\) is oriented from \(Q_3\) to \(Q_1\).

Since the origin region was to the left of the curves, it remains to the left, so the resulting region is \(\left\{{z{~\mathrel{\Big\vert}~}3\pi/4 < \operatorname{Arg}(z) < 5\pi/4}\right\}\):

From here, it’s a standard exercise, so to sum up:

• Rotate \(R\to \tilde R\) by \(z\mapsto iz\) to get a horizontal lune with intersection points \(\pm 1\).
• Send \(-1\to \infty, 1\to 0\) by \(z\mapsto {z+1\over z-1}\) to send \(\tilde R \to 3\pi/4 < \operatorname{Arg}(z) < 5\pi/4\).
• Reflect by \(z\mapsto -z\) to get \(-\pi/4 < \operatorname{Arg}(z) < \pi/4\).
• “Open” by \(z\mapsto z^{\pi \over 2 \theta_0}\) to map to \(-\pi/2<\operatorname{Arg}(z) < \pi/2\), where here \(\theta_0 \coloneqq{\pi \over 4}\).
• Rotate by \(z\mapsto iz\) to get \(0 < \operatorname{Arg}(z) < {\pi \over 2}\), i.e. \({\mathbb{H}}\).
• Use the Cayley map \(z\mapsto {z-i\over z+i}\) to send \({\mathbb{H}}\to {\mathbb{D}}\).

The main step: blow up the tangency.

The individual maps:

• Send \(0\to \infty\) by \(z\mapsto {1\over z}\) to map \(R\) to \(0<\Re(z)<1\)
• Rotate with \(z\mapsto iz\) to map this to \(0< \Im(z) < 1\)
• Dilate by \(z\mapsto \pi z\) to get \(0 < \Im(z) < \pi\)
• Apply \(z\mapsto e^z\) to map to \({\mathbb{H}}\).

That steps 1 and 2 work requires a bit of analysis. Use that \(f(z) \coloneqq 1/z\) satisfies \(f({\mathbb{R}}) = {\mathbb{R}}\) and \(f(i{\mathbb{R}}) = i {\mathbb{R}}\). To see where the circle \(C_1\) gets mapped to, parameterize it as \begin{align*} \gamma(t) \coloneqq\left\{{{1\over 2}\qty{1+e^{it}} {~\mathrel{\Big\vert}~}t\in [-\pi, \pi]}\right\} .\end{align*}

Now computing its image: \begin{align*} f(\gamma(t)) &= {2 \over 1 + e^{it}} \\ &= {2e^{-it\over 2} \over e^{-it\over 2}(1 + e^{it})} \\ &= {2e^{-it} \over e^{-it\over 2} + e^{it\over 2} } \\ &= { 2e^{-it\over 2} \over 2\cos\qty{t\over 2} } \\ &= \sec\qty{t\over 2}\qty{\cos\qty{-t\over 2} + i\sin\qty{-t\over 2}} \\ &= \sec\qty{t\over 2}\qty{\cos\qty{t\over 2} - i\sin\qty{t\over 2}} \\ &= 1 - i\tan\qty{t\over 2} ,\end{align*} and as \(t\) runs from \(-\pi\) to \(\pi\), \(-\tan\qty{t\over 2}\) runs from \(+\infty\) to \(-\infty\). So this is a line along \(z=1\), oriented top to bottom as in the image.

Similarly, computing \(f(i{\mathbb{R}})\): parameterize \(\gamma(t) = it\) with \(t\in (-\infty, \infty)\), then \begin{align*} f(\gamma(t)) = {1\over it} = -{i\over t} ,\end{align*} and as \(t\) runs from \(0\) to \(\infty\), \(-1/t\) runs from \(-\infty\) to \(0\), and as \(t\) runs from \(-\infty\) to \(0\), \(-1/t\) runs from \(0\) to \(\infty\). So \(f(i{\mathbb{R}})\) is oriented from bottom to top, as in the image.

That the region outside the disc is mapped to the strip shown: points \(x\in {\mathbb{R}}\) with \({\left\lvert {x} \right\rvert} > 1\) map to \({\left\lvert {x} \right\rvert}<1\), which is in the strip. One can also conclude this by handedness: the original region is on the right with respect to \(C_1\) and also on the right with respect to \(i{\mathbb{R}}\), so the new region should be on the right with respect to both \(f(i{\mathbb{R}})\) and \(f(C_1)\) with their induced orientations.

This is a lune with a single intersection vertex at \(z=i\). Orient the circles positively.

• Take \(f(z) = {z+i\over z-i}\) to send

  • \(i\to \infty\)
  • \(-i\to 0\)
  • \(1\to {1+i\over 1-i} = i\)
  • \(0\to -1\)

  So \({\left\lvert {z} \right\rvert} = 1\) is sent to the imaginary axis \(\left\{{it}\right\}\) for \(t\in (-\infty, \infty)\) oriented positively and \({\left\lvert {z- {i\over 2}} \right\rvert} = {1\over 2}\) is sent to \(\left\{{-1 + it}\right\}\) also oriented positively. The region then maps to \(-1 < \Re(z) < 0\).

• Rotate by \(z\mapsto -i\pi z\) to get \(0 < \Im(z) < \pi\).

• Take \(z\mapsto e^z\) to get \({\mathbb{H}}\).

• Take \(z\mapsto {z-i\over z+i}\) to get \({\mathbb{D}}\).

Note: this seems unusually difficult for a UGA question! This is a bigon with one vertex \(z_2 = 0\), so send it to infinity and keep track of the slit.

In steps:

• Use \(z\mapsto 1/z\) to get a vertical strip \(0<\Re(z) < 1\), where the slit maps to \([1/2, 1]\).

• Shift and dilate with \(z\mapsto \pi(z-1/2)\) to get \(-\pi/2<\Re(z) < \pi/2\), where the slit is now at \([0, \pi/2]\).
• Apply \(z\mapsto \sin(z)\) to get \({\mathbb{C}}\setminus[0, 1]\).
• Move the slit with \(z\mapsto 4(z-1/2)\) to get \({\mathbb{C}}\setminus[-2, 2]\).
• Apply \(z\mapsto z+z^{-1}\) to get \({\mathbb{C}}\setminus{\mathbb{D}}\), where the slit is now eliminated.
• Use \(z\mapsto 1/z\) to get \({\mathbb{D}}\)
• Use the inverse Cayley map \(z\mapsto i{1-z\over 1+z}\) to get \({\mathbb{H}}\).

The region looks like the following:

Following the general strategy for lunar regions, send the intersection points to \(0\) and \(\infty\) to get triangular sector. So choose to send \(0\to 0\) and \(1\to \infty\) by taking \begin{align*} f(z) \coloneqq{z\over z-1} .\end{align*}

Note: mistake here, really we need to compose with \(z\mapsto -z\) to get the picture, so take \(f(z) \coloneqq{z\over 1-z}\) instead!!

From here it is easy to map to the disc:

• \(z\mapsto {z\over z-1}\) sends \(R\) to \({\left\lvert {\operatorname{Arg}(z)} \right\rvert} < \theta_0\)
• \(z\mapsto z^{\pi \over 2\theta_0}\) maps \({\left\lvert {\operatorname{Arg}(z)} \right\rvert}<\theta_0 \to {\left\lvert {\operatorname{Arg}(z)} \right\rvert} < {\pi \over 2}\), the right half-plane.
• \(z\mapsto iz\) rotates the right half-plane into \({\mathbb{H}}\).
• \(z\mapsto {z-i\over z+i}\) maps \({\mathbb{H}}\to {\mathbb{D}}\).

Part a: That \(f: {\mathbb{C}}\setminus\left\{{0}\right\}\to {\mathbb{D}}\setminus\left\{{0}\right\}\) is injective: compute the derivative as \begin{align*} f'(z) = {1\over 2}\qty{1 - {1\over z^2}} ,\end{align*} which only vanishes at \(z=\pm 1\). Away from 0 in \({\mathbb{D}}\), \(f'\) is nonzero and continuous, so by the inverse function theorem \(f\) is a local homeomorphism onto its image, and in particular is injective.

The images of circles: parameterize one as \(\gamma(t) = Re^{it}\) for \(t\in [-\pi, \pi]\). Note that if \(R=1\), \(f(\gamma(t)) = {1\over 2}\qty{e^{it} + e^{-it}} = \cos(t)\), so as \(t\) increases from \(-\pi\to \pi\), the interval \([-1, 1]\) is covered twice. For \(0<R<1\), \begin{align*} f(\gamma(t)) &= {1\over 2}\qty{ Re^{it} + {1\over Re^{it}}} \\ &= {1\over 2}\qty{Re^{it} + R^{-1}e^{-it} } \\ &= {1\over 2}\qty{R\cos(t) + iR\sin(t) + R^{-1}\cos(-t) + iR^{-1}\sin(-t) } \\ &= {1\over 2}\qty{R\cos(t) + iR\sin(t) + R^{-1}\cos(t) - iR^{-1}\sin(-t) } \\ &= {1\over 2}\qty{R+R^{-1}}\cos(t) + i{1\over 2}\qty{R-R^{-1}}\sin(t) \\ &\coloneqq H_R \cos(t) + iV_R\sin(t) ,\end{align*} which is generally the equation of an ellipse of horizontal radius \(H_R\) and vertical radius \(V_R\). As \(R\) varies, these sweep out ellipses of vertical radii from 0 to \(\infty\). One can compute the foci: their distance from \(z=0\) is given by \(c\), where \begin{align*} c^2 = H_R^2 - V_R^2 = {1\over 4}(R+R^{-1})^2 - {1\over 4}(R-R^{-1})^2 = 1 ,\end{align*} so the foci are all at \(\pm 1\in {\mathbb{R}}\). One can check that these are clockwise when \(0<R<1\) and counterclockwise when \(R>1\).

In general, you take the coefficient for the major axis squared minus that of the minor axis squared. The foci are along the major axis.

Part b: The claim is that \(f({\mathbb{C}}\setminus{\mathbb{D}}) = {\mathbb{C}}\setminus[-1, 1]\).

Note that \(f(z) = f(1/z)\), so for \(z\neq 1/z\) there are exactly two preimages. These points are exactly \(z=\pm 1\), so we need to take the domain \(\Omega \coloneqq{\mathbb{C}}\setminus({\mathbb{D}}\cup\left\{{\pm 1}\right\})\) to get injectivity. Otherwise, for every \(z\in \Omega\), exactly one of \(z\) or \(1/z\) is in \({\mathbb{D}}\), so \(f(z)\) takes on unique values in \(\Omega\). By part 1, the images of circles of radius \(R\) are ellipses, and these sweep out the entire plane outside of \([-1, 1]\):

To be explicit, one can just solve for the two preimages. Setting \(w=f(z)\) and solving for \(z\) yields \begin{align*} w &= {1\over 2}\qty{z + {1\over z}} \\ \implies 2wz &= z^2 + 1 \\ \implies z^2 - 2wz + 1 &= 0 \\ \implies z &= {2w \pm \sqrt{4w^2 -4} \over 2} \\ \implies z &= w\pm \sqrt{w^2-1} ,\end{align*} where in order to define \(\sqrt{w^2-1}\), one needs a branch cut for \(\operatorname{Log}\) along \([-1, 1]\), which is precisely what we’re deleting from the image.

Part c: as in a usual conformal map problem, find a map to \({\mathbb{C}}\setminus[-1,1]\to {\mathbb{D}}\).

• Send \(-1\to 0\) and \(1\to \infty\) with \(z\mapsto {z+1\over z-1}\). Checking that \(f(0) = -1\), this yields \({\mathbb{C}}\setminus{\mathbb{R}}_{\leq 0}\).

• Unwrap with \(z\mapsto \sqrt{z}\) to obtain the right half-plane \(-\pi/2<\operatorname{Arg}(z) < \pi/2\).

• Apply the rotated Cayley map \(z\mapsto {z-1\over z+1}\) to map this to \({\mathbb{D}}\).

This composes to \begin{align*} g(z) &= {\sqrt{z+1\over z-1} -1 \over \sqrt{z+1\over z-1} +1} \\ &= { \qty{ \sqrt{z+1} - \sqrt{z-1}}^2 \over (z+1)-(z-1) } \\ &= z - \sqrt{z^2-1} ,\end{align*} and checking that it has the right inverse: \begin{align*} w &= z-\sqrt{z^2-1} \\ \implies (z-w)^2 &= z^2-1 \\ \implies w^2+1 &= 2wz \\ \implies z = {1\over 2}\qty{w + {1\over w}} .\end{align*}

Consider the images of arcs \(\gamma_R(t) \coloneqq Re^{it}\) for \(t\in [0, \pi]\) and \(0<R<1\), which fill out the upper half disc: \begin{align*} f(Re^{it}) = - {1\over 2}\qty{ \qty{R+R^{-1}}\cos(t) + i(R-R^{-1})\sin(t) } \coloneqq H_R\cos(t) + V_R\sin(t) ,\end{align*} where \(H_R \coloneqq-{1\over 2}{R+R^{-1}}\) and \(V_R \coloneqq-{1\over 2}(R-R^{-1})\). Note that \(H_R\in (-\infty, -1]\) and \(V_R \in (0, \infty)\) – in particular, since \(V_R>0\), as \(t\) ranges through \((0, \pi)\), this traces out the top half of an ellipse (noting that \(V_R<0\) would trace out the bottom).

As \(R\) ranges from \((0, 1)\), \(V_R\) ranges from \((0, \infty)\), so this traces out all such elliptic arcs in \({\mathbb{H}}\) passing through \(H_R <0, iV_R \in {\mathbb{H}}, -H_R>0\). So these trace out all of \({\mathbb{H}}\).

In steps:

• Unfold with \(z\mapsto z^2\) to get \({\mathbb{D}}\cap{\mathbb{H}}\).
• Joukowski it with \(z\mapsto -{1\over 2}(z+z^{-1})\) to get \({\mathbb{H}}\).
• Fold with \(z\mapsto z^{1\over 2}\) to get \(Q_1 = A\).

Main idea: both circles can be uniformized in ways that send \(z_2, z_2'\) to zero. Handling \(C\):

• \(f_1\): Uniformize by translating and scaling \(C\) to \(S^1\). Note that the image of \(z_1\) is in \(S^1\)
• \(f_2\): if \(z_2\) was not enclosed by \(C\), apply \(z\mapsto 1/z\) to move \(z_2\) into \({\mathbb{D}}\). Otherwise just take \(f_2 = \operatorname{id}\). In either case, \(z_1\in S^1\) is preserved.
• \(f_3\): apply a Blaschke factor \(\psi_a(z)\) to move \(z_2\to 0\) in \({\mathbb{D}}\). Since \(\psi_a\) preserves \(S^1\), \(z_1\) is moved to another point in \(S^1\).

Define \(F\coloneqq f_3 \circ f_2 \circ f_1\). Similarly define \(g_1,\cdots, g_3\) and \(G\) for \(C'\), so \(z_2'\to 0\) in \({\mathbb{D}}\). Now \(F(z_1), G(z_1')\in S^1\), so there is a rotation \(h: z\mapsto \lambda z\) that sends \(F(z_1)\to G(z_1')\).

Take the final map to be \(f\coloneqq G\circ h \circ F^{-1}\).

We have \(f: \Delta^* \to \left\{{\Re(z) > 0}\right\}\), so let \(T: \left\{{\Re(z) > 0}\right\}\to \Delta\) be the rotated Cayley map \(T(z) = {z-1\over z+1}\). Then \(G\coloneqq T^{-1}\circ f: \Delta^* \to \Delta\), and since \({\left\lvert {T} \right\rvert} < 1\), \(z=0\) is a removable singularity for \(F\) and \(F\) extends holomorphically to \(G:\Delta\to {\mathbb{C}}\), since a priori \(G(0)\) may not be bounded by 1. Supposing \(G\) is not constant, \(G(\Delta) \subseteq \bar{\Delta)\) by continuity and \(G(\Delta)\) is open by the open mapping theorem, and \(\Delta = f(\Delta) \subseteq G(\Delta)\), so \(G:\Delta\to \Delta\) is a map of the disc. Then \(F \coloneqq T\circ G: \Delta\to \left\{{\Re(z) > 0}\right\}\) is an extension of \(F\) which is bounded in neighborhoods of \(z=0\), making zero a removable singularity for \(f\).

Write \(T:{\mathbb{C}}\to {\mathbb{D}}\) for the Cayley map, then \(F\coloneqq f\circ T\) satisfies \(F({\mathbb{C}}) = T(f({\mathbb{C}})) \subseteq T({\mathbb{H}}) = {\mathbb{D}}\), so \(F\) is a bounded entire function and thus constant. So \(c = F(z) = T(f(z)) \implies f(z) = T^{-1}(c)\), making \(f\) constant.

Part 1:

Write \(\psi_a(z) \coloneqq{a-z\over 1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu z}\) for the Blaschke factor of \(a\), and define \begin{align*} g(z) \coloneqq{f(z) \over \psi_a(z) \psi_{-a}(z)} .\end{align*}

In particular, \({\left\lvert {g(0)} \right\rvert} \leq 1\), so \begin{align*} 1\geq {\left\lvert {g(0)} \right\rvert} = {{\left\lvert {f(0)} \right\rvert} \over {\left\lvert {B_a(0)} \right\rvert} \cdot {\left\lvert {B_{-a}(0)} \right\rvert}} = {{\left\lvert {f(0)} \right\rvert} \over {\left\lvert {a} \right\rvert}^2} \implies {\left\lvert {a} \right\rvert}^2 \geq {\left\lvert {f(0)} \right\rvert} .\end{align*}

Part 2: Applying Schwarz-Pick: \begin{align*} {\left\lvert {f'(0)} \right\rvert} \leq {1 - {\left\lvert {f(0)} \right\rvert}^2 \over 1 - {\left\lvert {0} \right\rvert}^2 } = 1-{\left\lvert {a} \right\rvert}^2 < 1 ,\end{align*} using that \(a\neq 0\), so \(f\) is a contraction.

Can write \(f_e(z) \coloneqq{f(a) + f(-a) \over 2}\) to write \(f_e(z) = g(z^2)\). Compose with some \(\psi_a\) to get \(0\to 0\) and apply Schwarz – unclear how to unwind what happens in the case of equality though.

First show the hint: assume \(f\) has nonzero zeros. Write \(Z(f) \coloneqq f^{-1}(0)\) for the set of zeros in \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu\).

So write \(Z(f) = \left\{{z_1,\cdots, z_m}\right\}\) and define \begin{align*} g(z) \coloneqq\prod_{1\leq k \leq m} {z-z_k \over 1 - \mkern 1.5mu\overline{\mkern-1.5muz_k\mkern-1.5mu}\mkern 1.5mu z}, \quad h(z) \coloneqq{f(z) \over g(z)} .\end{align*}

So we now have \begin{align*} f(z) = e^{i\theta} \prod_{1\leq k\leq m} {z-z_k \over 1 -\mkern 1.5mu\overline{\mkern-1.5muz_k\mkern-1.5mu}\mkern 1.5mu z} ,\end{align*} which has poles at points \(z\) for which \(\mkern 1.5mu\overline{\mkern-1.5muz_k\mkern-1.5mu}\mkern 1.5muz=1\) for some \(z_k\in Z(f)\). However, since we assumed \(f\) was entire, it can have no such poles, which forces \(z_k = 0\) for all \(k\). But then \begin{align*} f(z) = e^{i\theta}\prod_{1\leq k \leq m}{z- 0 \over 1 - 0\cdot z} = e^{i\theta}z^m .\end{align*}

Proof due to Swaroop Hegde!

Fix \(R, M\) and make a clever choice: define \begin{align*} F: {\mathbb{D}}&\to {\mathbb{C}}\\ z &\mapsto {f(Rz) \over M} .\end{align*} Write \(a\coloneqq F(0)\) and consider the Blaschke factor \begin{align*} \psi_a(z) \coloneqq{a-z \over 1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu z} \in \mathop{\mathrm{Aut}}({\mathbb{D}}) ,\end{align*} and define \begin{align*} g: {\mathbb{D}}&\to {\mathbb{D}}\\ z &\mapsto (\psi_a \circ F)(z) .\end{align*} Then \(g(0) = 0\) and \({\left\lvert {g(z)} \right\rvert} \leq 1\) for all \(z\in {\mathbb{D}}\), so by Schwarz we have \({\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\) for all \(z\in {\mathbb{D}}\). Thus for all \(z\in {\mathbb{D}}\), \begin{align*} &{\left\lvert {g(z)} \right\rvert} \leq z \\ \\ \iff & {\left\lvert {\psi_a(F(z)) } \right\rvert} \leq {\left\lvert {z} \right\rvert} \\ \\ \iff & {\left\lvert { {f(Rz) \over M} - a \over 1 - {\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu f(Rz) \over M} } \right\rvert} \leq {\left\lvert {z} \right\rvert} \\ \\ \iff & {\left\lvert {f(Rz) - f(0) \over 1 - {\mkern 1.5mu\overline{\mkern-1.5muf(0)\mkern-1.5mu}\mkern 1.5mu f(Rz) \over M^2 } } \right\rvert} \leq {\left\lvert {z} \right\rvert} \\ \\ \iff & {\left\lvert {f(Rz) - f(0) \over M^2 - \mkern 1.5mu\overline{\mkern-1.5muf(0)\mkern-1.5mu}\mkern 1.5mu f(Rz) } \right\rvert} \leq {{\left\lvert {z} \right\rvert} \over M} \\ \\ \iff & {\left\lvert {f(w) - f(0) \over M^2 - \mkern 1.5mu\overline{\mkern-1.5muf(0)\mkern-1.5mu}\mkern 1.5mu f(w) } \right\rvert} \leq {{\left\lvert {w} \right\rvert} \over MR} ,\end{align*} which holds for all \(w\in {\mathbb{D}}\) by replacing \(Rz\) with \(w\) (i.e. to show this equality for arbitrary \(w\in {\mathbb{D}}\), write \(w = Rz\) for some \(z\in {\mathbb{D}}\) and run this chain of inequalities backward).

Set \begin{align*} g(z) \coloneqq{f(Rz+a) - f(a) \over 2M} ,\end{align*} so that \(g(0) = 0\) and \(g:{\mathbb{D}}\to {\mathbb{D}}\) so Schwarz applies, \begin{align*} {\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert} \implies {\left\lvert { f(Rz+a) - f(a) \over 2M } \right\rvert} &\leq {\left\lvert {z} \right\rvert} \\ \implies {\left\lvert { f(Rz+a) - f(a) } \right\rvert} &\leq 2M {\left\lvert {z} \right\rvert} \\ \implies {\left\lvert { f(w) - f(a) } \right\rvert} &\leq 2M{\left\lvert { w-a\over R} \right\rvert} \\ \implies {\left\lvert {f(w) - f(a) \over w-a} \right\rvert} &\leq {2M \over R} .\end{align*}

Define \(g:{\mathbb{H}}\to {\mathbb{H}}\) by \(g(z) = {1\over 2}iz\), so \(g(2) = i\). Then set \(F \coloneqq C\circ g \circ f: {\mathbb{D}}\to {\mathbb{D}}\) where \(C(z) \coloneqq{z-i\over z+i}\) is the Cayley map.Since \(F(0) = C(g(f(0))) = C(g(2)) = C(i) = 0\), Schwarz applies to \(F\) and \({\left\lvert {F'(z)} \right\rvert}\leq 1\) for \(z\in {\mathbb{D}}\). By the chain rule, \begin{align*} F'(z) = f'( (g\circ C) (z))\cdot g'(C(z)) \cdot C'(z) .\end{align*} Setting \(g(C(z)) = 0\) yields \(z=C^{-1}(g^{-1}(0)) = C^{-1}(0) = i\). \begin{align*} F'(i) &= f'(0) \cdot g'(0) \cdot C'(i) \\ \implies {\left\lvert {f'(0)} \right\rvert} &\leq {\left\lvert {F'(i) \over g'(0) C'(i)} \right\rvert} \\ &\leq {1\over {\left\lvert {g'(0)} \right\rvert} \cdot {\left\lvert {C'(i)} \right\rvert} } \\ &= {1\over {\left\lvert {i\over 2} \right\rvert} \cdot {\left\lvert {-{i\over 2} } \right\rvert} } \\ &= 4 .\end{align*}

By Schwarz, if \({\left\lvert {F'(z)} \right\rvert} = 1\) for any \(z\in {\mathbb{D}}\), we’ll have \(F(z) = \lambda z\) for some \({\left\lvert { \lambda} \right\rvert} = 1\). Unwinding this: \begin{align*} F(z) &= \lambda z \implies (C\circ g\circ f)(z) = \lambda z \\ \implies f(z) &= g^{-1}(C^{-1}(\lambda z)) = g^{-1}\qty{-i {\lambda z + 1 \over \lambda z - 1}} \\ \implies f(z) &= -2 {\lambda z + 1\over \lambda z - 1} .\end{align*} Moreover \(f'(z) = -2\qty{-2\lambda \over (\lambda z - 1)^2}\), so \begin{align*} {\left\lvert {f'(0)} \right\rvert} = 4{\left\lvert {\lambda} \right\rvert} = 4 .\end{align*}

Note \({\left\lvert {f(z)} \right\rvert}\leq 1\), and \(g\coloneqq f(z)/z^k\) has a removable singularity at zero since \(g\) is bounded on \({\mathbb{D}}\): fixing \({\left\lvert {z} \right\rvert} = r < 1\), \begin{align*} {\left\lvert {g(z)} \right\rvert} = {\left\lvert {f(z)\over z^k} \right\rvert} = {\left\lvert {f(z)} \right\rvert}r^{-k}\leq r^{-k}\overset{r\to 1}\longrightarrow 1 .\end{align*} So \(g:{\mathbb{D}}\to {\mathbb{D}}\) since \({\left\lvert {g(z)} \right\rvert}\leq 1\) on \({\mathbb{D}}\) by the MMP. Since \(g\) has no zeros on \({\mathbb{D}}\), by the MMP \({\left\lvert {g} \right\rvert} \geq 1\) on \({\mathbb{D}}\), so \({\left\lvert {g} \right\rvert} = 1\) is constant, making \(g(z) = \lambda z\) a rotation. Then \(f(z) = \lambda z^n\).

Alternative to MMP: if \(g\) has no zeros in \({\mathbb{D}}\), \(g\) admits a conjugate reflection through \({\mathbb{D}}\) by \(z\mapsto 1/\mkern 1.5mu\overline{\mkern-1.5muf(1/\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu\). This is bounded and entire, thus constant, making \(g\) constant.

Since \(f\) is injective, it has a left-inverse \(f^{-1}\), and \(F\coloneqq f^{-1}g\) is well-defined. Since \(F:{\mathbb{D}}\to {\mathbb{D}}\) and \(F(0) = 0\), Schwarz applies and \({\left\lvert {F(z)} \right\rvert} \leq z\) on \({\mathbb{D}}\). Unwinding: \begin{align*} {\left\lvert {(f^{-1}\circ g)(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert} \implies {\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {f(z)} \right\rvert} \qquad \forall {\mathbb{D}}\in {\mathbb{Z}} .\end{align*} This says that \(g({\mathbb{D}}) \subseteq f({\mathbb{D}})\), and in particular this holds on all \({\mathbb{D}}_r(0)\), so \(g({\mathbb{D}}_r(0)) \subseteq f({\mathbb{D}}_r(0))\).

Define a function \begin{align*} F(z) \coloneqq \begin{cases} f(z) & \Im(z)\geq 0 \\ \mkern 1.5mu\overline{\mkern-1.5muf(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu & \Im(z) < 0 \end{cases} .\end{align*} Then \(F\) is holomorphic on \(\tilde S\coloneqq\left\{{z\in {\mathbb{D}}{~\mathrel{\Big\vert}~}\Im(z) < 0}\right\}\) – write \(w_0\in \tilde S\) as \(w_0 = \mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu\) for some \(z_0\in S\), then \begin{align*} f(z) &= \sum_{k\geq 0}c_k (z-z_0)^k \\ \implies \mkern 1.5mu\overline{\mkern-1.5muf(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu &= \sum_{k\geq 0}\mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu \mkern 1.5mu\overline{\mkern-1.5mu\qty{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - z_0}^k\mkern-1.5mu}\mkern 1.5mu \\ &= \sum_{k\geq 0}\mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu \qty{z - \mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu}^k \\ &= \sum_{k\geq 0}\mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu \qty{z - w_0}^k ,\end{align*} which yields a power series expansion of \(F\) about \(w_0\). So \(f\) is analytic at every point in \(\tilde S\) and thus holomorphic. Since \(\mkern 1.5mu\overline{\mkern-1.5muf(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu = f(z)\) for \(z, f(z)\in {\mathbb{R}}\), \(F\) is a continuous extension of \(f\) to \({\mathbb{D}}\). By the symmetry principle, \(F\) is holomorphic, and \({ \left.{{F}} \right|_{{S}} } = f\).

This is the Schwarz–Pick lemma.

• Fix \(z_1\) and let \(w_1 = f(z_1)\). Define \begin{align*} \psi_{a}(z) \coloneqq{a-z \over 1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5muz} \in \mathop{\mathrm{Aut}}({\mathbb{D}}) .\end{align*}

  • Note that inequality now reads \begin{align*} {\left\lvert {\psi_{f(w)}(f(z)) } \right\rvert} \leq {\left\lvert {\psi_w(z)} \right\rvert} .\end{align*} Moreover \(\psi_a\) is an involution that swaps \(a\) and \(0\).

• Now set up a situation where Schwarz’s lemma will apply: \begin{align*} 0 \xrightarrow{\psi_{z_1}} z_1 \xrightarrow{f} f(z) \xrightarrow{\psi_{f(z_1)}} 0 ,\end{align*} so \(F\coloneqq\psi_{f(z_1)} \circ f \circ \psi_{z_1} \in \mathop{\mathrm{Aut}}({\mathbb{D}})\) and \(F(0) = 0\).

• Apply Schwarz we get \({\left\lvert {F(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}\) for all \(z\), so \begin{align*} {\left\lvert {F(z)} \right\rvert} &\leq {\left\lvert {z} \right\rvert} \\ \implies {\left\lvert { f(z_1) - (f\circ \psi_{z_1})(z) \over 1 - \mkern 1.5mu\overline{\mkern-1.5muf(z_1)\mkern-1.5mu}\mkern 1.5mu \cdot (f\circ \psi_{z_1}) (z) } \right\rvert} &\leq {\left\lvert { z} \right\rvert} \\ \implies {\left\lvert {f(z_1) - f(w) \over 1 - \mkern 1.5mu\overline{\mkern-1.5muf(z_1)\mkern-1.5mu}\mkern 1.5mu\cdot f(w) } \right\rvert} &\leq {\left\lvert {\psi_{z_1}(z)} \right\rvert} && w\coloneqq\psi_{z_1}(z) \\ \implies {\left\lvert {f(z_1) - f(w) \over 1 - \mkern 1.5mu\overline{\mkern-1.5muf(z_1)\mkern-1.5mu}\mkern 1.5mu\cdot f(w) } \right\rvert} &\leq {\left\lvert {z_1 - z \over 1 - \mkern 1.5mu\overline{\mkern-1.5muz_1\mkern-1.5mu}\mkern 1.5mu z } \right\rvert} .\end{align*}

• Since \(z_1\) was arbitrary and fixed and \(w\) was a free variable, this holds for all \(z,w\in {\mathbb{D}}\).

• Strictness: suppose equality holds, we’ll show that \(f(z) = {az+b\over cz+d}\)

• By Schwarz, \(F(z) = \lambda z\) for \(\lambda \in S^1\). Thus \begin{align*} (\psi_{f(z_1)} \circ f \circ \psi_{z_1}) (z) &= \lambda z \\ \implies (f \circ \psi_{z_1}) (z) &= \psi_{f(z_1)}^{-1}(\lambda z ) \\ \implies f(w) &= \psi_{f(z_1)}^{-1}(\lambda \psi_{z_1}^{-1}(w) ) && w\coloneqq\psi_{z_1}(z) \\ &= \psi_{f(z_1)} \qty{\lambda \psi_{z_1}(w)} \\ &= \lambda \psi_{\mkern 1.5mu\overline{\mkern-1.5mu\lambda \mkern-1.5mu}\mkern 1.5muf(z_1)} \qty{\psi_{z_1}(w)} \\ &\coloneqq\lambda \psi_a(\psi_b(w)) \\ &=\lambda\qty{ a- \psi_b(w) \over 1 - \mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu \psi_b(w) } \\ &= \quad \vdots \\ &= -\lambda \qty{ \frac{{\left(a \overline{b} - 1\right)} z - a + b}{{\left(\overline{a} - \overline{b}\right)}z - b \overline{a} + 1} } \\ &= \qty{ \frac{-\lambda {\left(a \overline{b} - 1\right)} z + \lambda( a - b)}{{\left(\overline{a} - \overline{b}\right)}z + (- b \overline{a} + 1)} } ,\end{align*} which is evidently a linear fractional transformation.

Given this, there’s just a clever rearrangement to obtain the stated result: \begin{align*} {\left\lvert {f(w) - f(z) \over 1 - \mkern 1.5mu\overline{\mkern-1.5muf(w)\mkern-1.5mu}\mkern 1.5muf(z)} \right\rvert} &\leq {\left\lvert {w-z \over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5mu z} \right\rvert} \\ \implies {\left\lvert { 1\over 1-\mkern 1.5mu\overline{\mkern-1.5muf(w)\mkern-1.5mu}\mkern 1.5muf(z) } \right\rvert} \cdot {\left\lvert {f(z) - f(w) \over z-w} \right\rvert} &\leq {\left\lvert {1\over 1-\mkern 1.5mu\overline{\mkern-1.5muw\mkern-1.5mu}\mkern 1.5muz} \right\rvert} \\ ,\end{align*} and taking \(z\to w\) on both sides yields \begin{align*} {\left\lvert {1\over 1 - {\left\lvert {f(w)} \right\rvert}^2 } \right\rvert} {\left\lvert {f'(w)} \right\rvert} \leq {1\over {\left\lvert {w} \right\rvert}^2} \implies {\left\lvert {f'(w)} \right\rvert} \leq {1-{\left\lvert {f(w)} \right\rvert}^2\over 1-{\left\lvert {w} \right\rvert}^2 } .\end{align*}

Part 1: use the reflection principle to define \begin{align*} F(z) \coloneqq \begin{cases} f(z) & {\left\lvert {z} \right\rvert} \leq 1 \\ {1\over \mkern 1.5mu\overline{\mkern-1.5muf\qty{1/\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\mkern-1.5mu}\mkern 1.5mu } & {\left\lvert {z} \right\rvert} \geq 1 \end{cases} .\end{align*}

Now \(F:{\mathbb{CP}}^1\to {\mathbb{CP}}^1\) is holomorphic and all such functions are rational. As a consequence, \(f\) is rational.

Part 2: As in the proof of Schwarz, define \(g(z) \coloneqq{f(z)\over z^n}\) where \(n = {\operatorname{Ord}}_{f}(0)\). Then \(g\) is holomorphic on \({\mathbb{D}}\) since the singularity at \(z=0\) is removable. On \({\left\lvert {z} \right\rvert} = r<1\), \begin{align*} {\left\lvert {g(z)} \right\rvert} = { {\left\lvert {f(z)} \right\rvert} \over {\left\lvert {z} \right\rvert} } = {{\left\lvert {f(z)} \right\rvert} \over r} \leq {1\over r} \overset{r\to 1^-}\longrightarrow 1 ,\end{align*} using that \({\left\lvert {f} \right\rvert} \leq 1\) on \({\mathbb{D}}\). By the MMP, \({\left\lvert {g} \right\rvert} \leq 1\) on all of \({\mathbb{D}}\). Note that \({\left\lvert {g} \right\rvert} = 1\) when \({\left\lvert {z} \right\rvert}=1\), so \({\left\lvert {1/g} \right\rvert}\leq 1\) in \({\mathbb{D}}\) by the MMP, forcing \({\left\lvert {g} \right\rvert} = 1\). Unwinding this, \({\left\lvert {f} \right\rvert} = {\left\lvert {z} \right\rvert}^n\), go \(f(z) = \lambda z^n\) for some \({\left\lvert {\lambda} \right\rvert} = 1\).

Part 3: Define \(\Psi(z) \coloneqq\prod_{k\leq n} \psi_{a_k}(z)\) where \(\psi_a(z) \coloneqq{a-z\over 1-\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu z}\). Set \(g(z) \coloneqq{f(z) \over \Psi(z)}\), then by the same argument as above, \({\left\lvert {g} \right\rvert} \leq 1\) and \({\left\lvert {g} \right\rvert} = 1\) on \({\left\lvert {z} \right\rvert} = 1\). Then \(g\) has no zeros, since they’ve all been divided out, and no poles since \(f\) is holomorphic on \({\mathbb{D}}\), so \(1/g\) is holomorphic on \({\mathbb{D}}\). Since \({\left\lvert {1/g} \right\rvert} = 1\) on \(S^1\), this forces \(g\) to be constant. Equality in the Schwarz lemma implies \(g(z) = \lambda z\) is a rotation, and unwinding this yields \(f(z) = \lambda \Psi(z)\).

Use Rouché: if \({\left\lvert {f(z)} \right\rvert} < 1\) is strict when \({\left\lvert {z} \right\rvert} = 1\), then consider \(F(z) \coloneqq f(z) - z\). Write the big part as \(M(z) = z\) and the small as \(m(z) = f(z)\), then on \({\left\lvert {z} \right\rvert} = 1\) \begin{align*} {\left\lvert {m(z)} \right\rvert} = {\left\lvert {f(z)} \right\rvert} < 1 = {\left\lvert {z} \right\rvert} = {\left\lvert {M(z)} \right\rvert} ,\end{align*} so \(M(z)\) and \(m(z) + M(z) = f(z) - z\) have the same number of zeros in \({\mathbb{D}}\) – precisely one.

There is still a unique fixed point. Use the Brouwer fixed point theorem: since \(g\) is holomorphic on \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu\), it is in particular continuous. By the Brouwer fixed point theorem, every continuous map \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu \to \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu\) has a fixed point. If \(g\) is nonconstant, then the fixed point is unique by Schwarz: without loss of generality one can assume \(f(0) = 0\) by composing with a Blaschke factor. Apply Schwarz to \(f\), then if \(f(a) = a\) we have the equality clause and \(f(z) = \lambda z\). Since \(a = f(a) = \lambda a\), \(\lambda = 1\) and \(f\) is the identity. If \(g\) is constant, then \({\left\lvert {g(z)} \right\rvert} < 1\) on \({\left\lvert {z} \right\rvert} = 1\) forces \(g\equiv 0\).

Note that there is a major difference between self maps to \({\mathbb{D}}\) versus \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu\). By the argument in part 2, if \(f(z)\) is not the identity then \(f\) can have at most one fixed point. Moreover, not every map \(f:{\mathbb{D}}\to{\mathbb{D}}\) need have a fixed point: consider \begin{align*} g: {\mathbb{H}}&\to {\mathbb{H}}\\ z &\mapsto z+1 .\end{align*} Now conjugate with the Cayley map \(C:{\mathbb{H}}\to {\mathbb{D}}\) to define \(f\coloneqq CgC^{-1}:{\mathbb{D}}\to {\mathbb{D}}\) which has no fixed points at all.

Define \(F(z) \coloneqq{f(z) \over g(z)}\).

Given this, note that \({\left\lvert {F(z)} \right\rvert} = 1\) for all \(z\), so \(F(\Omega) \subseteq S^1\), which is codimension 1 in \({\mathbb{C}}\) and not open. By the open mapping theorem, \(F\) must be constant, so \(F(z) = \lambda\), and in particular since \({\left\lvert {F(z)} \right\rvert} = 1\), \(\lambda = e^{it}\in S^1\) for some \(t\). Then \(f(z) = \lambda g(z)\).

Note that there is a unique fixed point. We have \(f(0) = 0\), so there is at least one, so suppose \(a\) is another fixed point with \(f(a) = a\). By Schwarz, \({\left\lvert {f(z)} \right\rvert}\leq {\left\lvert {z} \right\rvert}\) with equality at any nonzero point implying \(f\) is a rotation, and \(f(a) = a\implies {\left\lvert {f(a)} \right\rvert} = {\left\lvert {a} \right\rvert}\), so write \(f(z) = e^{i\theta}z\). Now \(f(a) = a = e^{i\theta }a\) forces \(\theta = 0\), so \(f(z) = z\) is the identity.

Since \(f(0) = 0\), the Schwarz lemma applies and either

• \(f(z) = e^{i\theta} z\) is a rotation, or
• \({\left\lvert {f'(0)} \right\rvert} < 1\) and \({\left\lvert {f(z)} \right\rvert} < z\) for all \(z\in {\mathbb{D}}\).

Supposing the latter, \(f\) is a contraction, and \({\left\lvert {f_{n+1}(z)} \right\rvert} < {\left\lvert {f_{n}(z)} \right\rvert}\) for all \(n\) and all \(z\), so \({\left\lvert {f_n(z)} \right\rvert} \overset{n\to\infty}\longrightarrow 0\) for all \(z\). Since \(f_n\to g\) pointwise, this means \(g(z) = 0\) for all \(z\), making \(g\equiv 0\).

Otherwise, suppose \(f\) is a rotation. Then if \(f(z) = e^{i\theta}z\), \(f_n(z) = e^{in\theta}z\). The pointwise limit \(\lim_{n\to\infty}e^{in\theta}z\) can only exist if \(\theta = 0\), otherwise this is periodic when \(\theta\) is rational or the points \(e^{i\theta}z, e^{2i\theta }z,\cdots\) form form a countably infinite set of distinct points. So \(f(z) = z\), making \(\lim_{n\to \infty}f_n(z) = z\) as well.

Idea:

Let \(f: U\to {\mathbb{C}}\). Pick \(w_0\in W\) with \(f(z_0) = w_0\) for some \(z_0\in U\); we want to show that \(w_0\) is an interior point of \(f(U)\), so we’re looking for a disc containing \(w_0\) and contained in \(f(U)\).

Write \begin{align*} g_0(z) \coloneqq f(z) - w_0 ,\end{align*} so \(g_0\) is holomorphic and has a zero at \(z_0\). Since zeros of holomorphic functions are isolated, there is some \(U' \coloneqq{\mathbb{D}}_r(z_0)\) where \(g_0\) is nonvanishing. The claim is that if we choose \({\varepsilon}\) small enough, we can arrange so that \(W_{\varepsilon}\coloneqq{\mathbb{D}}_{\varepsilon}(w_0) \subseteq f(U)\). This will follow if for every \(w\in W_{\varepsilon}\), the equation \(f(z) = w\) has a solution in \(U\), i.e.  Define a function that counts the number of zeros: \begin{align*} F(w) &\coloneqq{1\over 2\pi i}\int_{{{\partial}}U' } {f(z) \over f(z) - w_1 }\,dz\\ &= {1\over 2\pi i}\int_{{{\partial}}U' } {{\frac{\partial }{\partial z}\,}\qty{f(z) - w} \over f(z) - w }\,dz\\ &= {\sharp}Z(f(z) - w, U' ) ,\end{align*} which is the number of zeros of \(f(z) - w\) in \(U'\) by the argument principle. Now \(F\) is a \({\mathbb{Z}}{\hbox{-}}\)valued function, and the only obstruction to continuity is if \(f(z) - w = 0\) in the integrand for some \(z\). The claim is that \({\varepsilon}\) can be chosen such \begin{align*} z\in {{\partial}}U' \implies {\left\lvert {f(z) - w} \right\rvert} > 0 \qquad \forall w\in W_{\varepsilon} .\end{align*} The theorem then follows immediately: \(F(w): U' \to W_{\varepsilon}\) is a continuous and \({\mathbb{Z}}{\hbox{-}}\)valued, thus constant. Then noting that \(F(w_0) = 1\) since \(z_0\in U'\) and \(w_0\in W_{\varepsilon}\), we have \(F\equiv 1 > 0\) for all \(w\).

Note that \(f\) is non-constant, since a constant function is extremely non-injective. Consider the singularity at \(\infty\):

• If it is removable, then \(f\) is bounded outside of a large disc, and bounded inside of it as a continuous function on a compact set, making \(f\) entire and bounded and thus constant by Liouville.

• If it is essential, then by Casorati-Weierstrass there is a large disc of radius \(R\) such that \(f(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}_R\mkern-1.5mu}\mkern 1.5mu^c) \subseteq {\mathbb{C}}\) is dense. By the open mapping theorem, \(f({\mathbb{D}}_R) \subseteq {\mathbb{C}}\) is open, so by density it intersects \(f(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}_R\mkern-1.5mu}\mkern 1.5mu^c)\), but \({\mathbb{D}}_R \cap\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}_R\mkern-1.5mu}\mkern 1.5mu^c\) is empty so this contradicts injectivity.

So we can conclude \(\infty\) is a pole of some order \(N\), so \(f\qty{1\over z} = \sum_{0\leq k\leq N} c_k z^{-k}\) and thus \(f(z) = \sum_{0\leq k\leq N} c_k z^k\) is a polynomial of degree \(N\). However, a polynomial of degree \(N\) is generically \(N\)-to-one locally, so injectivity forces \(N=1\) and \(f(z) = c_0 + c_1 z\), where \(c_1\neq 0\) since \(f\) is nonconstant.

Write \(g(z) \coloneqq f(1/z)\), which has a singularity at \(z=0\). The claim is that this is a pole.

If \(z=0\) is a removable singularity, \(g\) is bounded on some closed disc \({\left\lvert {z} \right\rvert} \leq {\varepsilon}\), so \(f\) is bounded on \({\left\lvert {z} \right\rvert} > {\varepsilon}\). Moreover \(f\) is continuous and \({\left\lvert {z} \right\rvert}\leq {\varepsilon}\), \(f\) is bounded on this disc. This makes \(f\) an entire bounded function and thus constant by Liouville, contradicting injectivity.

If \(z=0\) is essential, then by Casorati-Weierstrass pick a punctured disc \(D = \left\{{{\left\lvert {z} \right\rvert} \leq {\varepsilon}}\right\}\) where \(g(D)\) is dense in \({\mathbb{C}}\). Writing \(D^c \coloneqq\left\{{{\left\lvert {z} \right\rvert} > {\varepsilon}}\right\}\), this means that \(f(D^c)\) is dense. But \(U\coloneqq\left\{{{\left\lvert {z} \right\rvert} < {\varepsilon}}\right\}\) is open and by the open mapping theorem \(f(U)\) is open, so by density there is a point \(w\in f(D^c) \cap f(U)\) while \(U \cap D^c = \emptyset\), again contradicting injectivity.

So \(z=0\) is a pole of \(g\), and \(g\) admits a Laurent expansion \begin{align*} g(z) = \sum_{k\geq -N} c_k z^k .\end{align*} Since \(f\) is entire, it equals its Laurent expansion at \(z=0\), so equating the two series yields \begin{align*} f(z) = \sum_{k\geq 0} d_k z^k &\implies g(z) = \sum_{k\geq 0} {d_k \over z^k} = \sum_{1\leq k\leq N} {c_k\over z^k} + \sum_{k\geq 0} c_k z^k \\ &\implies \sum_{k\geq 0} c_k z^k = 0 \\ &\implies f(z) = \sum_{0\leq k \leq N} c_k z^k ,\end{align*} making \(f\) a polynomial of degree at most \(N\).

Now \(f\) can not be degree zero, since constant maps are not injective. Moreover \(f\) can not be degree \(N\geq 2\), since any polynomial of degree \(N\) has \(N\) roots in \({\mathbb{C}}\) by the fundamental theorem of algebra, and any two distinct roots will be points where injectivity fails. Finally, ruling out the case of roots with multiplicity, if \(f(z) = c(z-a)^N\), then \(f\) has exactly \(N\) preimages in a neighborhood of \(a\). Letting \(p\) be any such point, we can find \(N\) complex points mapping to it: \begin{align*} p = c(z-a)^N &\implies {p\over c} = (z-a)^N \\ &\implies \qty{p\over c}^{1\over N}\zeta_N^k = z-a \quad k=0,1,\cdots, n-1 \\ &\implies z_k\coloneqq\qty{p\over c}^{1\over N}\zeta_N^k + a \xrightarrow{f} p .\end{align*}

So \(f\) must be degree exactly 1, i.e. \(f(z) = az+b\).

As in the proof of Casorati-Weierstrass, fix \(w\) and suppose toward a contradiction that no sequence sequence exists. Then there is some \({\varepsilon}, R\) such that \begin{align*} f({\mathbb{D}}_{\varepsilon}(a)) \subseteq {\mathbb{D}}_R(w)^c ,\end{align*} for otherwise one could construct the desired sequence. In particular, \({\left\lvert {f(z) - w} \right\rvert} > R\) for \({\left\lvert {z-a} \right\rvert} < {\varepsilon}\), so define \begin{align*} G(z) \coloneqq{1\over f(z) - w} \implies {\left\lvert {G(z)} \right\rvert} \leq R^{-1}< \infty \qquad \text{in }{\mathbb{D}}_{\varepsilon}(a) .\end{align*} Since \(G\) is bounded in this disc, any singularities here must be removable. Since the \(a_k\) are poles of \(f\), they are zeros of \(G\) – this is because if \({\left\lvert {f(z)} \right\rvert}\to\infty\) as \(z\to a_k\) then \({\left\lvert {G(z)} \right\rvert}\to 0\). So \(G(a_k) = 0\) for all \(k\) and \(G\) extends holomorphically over the removable singularity \(a\), and by continuity must satisfies \(G(a) = 0\). But now \(G\) is zero on a set with a limit point, hence \(G\equiv 0\) by the identity principle. This is a contradiction since if \(G\equiv 0\) on an open set, \(f\) has poles on an open set, contradicting that \(f\) is holomorphic on \(\Omega\).

The difference to Casorati-Weierstrass: the singularity at \(a\) is not essential, since in particular it is not isolated. The conclusion is nearly the same though: this says that every \(w\in {\mathbb{C}}\) is a limit point for \(f(\Omega)\), so \(w\) is in the closure of \(f(\Omega)\), making the image dense in \({\mathbb{C}}\).

Ideas:

• If an entire function doesn’t have dense image, it’s constant by Liouville using the proof idea of Casorati-Weierstrass.
• Conjugate \(f\) by \(T:{\mathbb{H}}\to {\mathbb{D}}\) where \(T(z) = {z-i\over z+i}\), then \(\tilde f(0) = 0\)
• Use that \(T({\mathbb{R}}) = S^1\), so \({\left\lvert {\tilde f(z)} \right\rvert} = 1\) when \({\left\lvert {z} \right\rvert} = 1\).
• Schwarz reflection applies to \(\tilde f\) to define an entire function – if \(f\) isn’t dense, then the extension of \(\tilde f\) isn’t dense…?
• No clue how to use \(f(i) = i\), although it implies \(\tilde f(0) = 0\) and Schwarz applies.

Part 1: This is not true: take the holomorphic function \(f(z) = z\), then \(u(z) \coloneqq\Re(f(z)) = \Re(z)\) is harmonic on nonzero on \({\mathbb{R}}\) but zero on \(i{\mathbb{R}}\).

Part 2: Set \(f \coloneqq u_x + i u_y\), then \(f\) is holomorphic on \(O\). Since \(h\equiv 0\) on \({\mathbb{D}}_{\varepsilon}\subseteq O\), \(g\equiv 0\) on this disc. By the identity principle for holomorphic functions, \(g\equiv 0\) on \(O\). So \(h_x, h_y \equiv 0\), making \(h\) constant, and since \(h\equiv 0\) on \(U\) this forces \(h\equiv 0\) on \(O\).

Part 3: Let \(u\) be harmonic on \(S^+\), a region symmetric about \({\mathbb{R}}\), and that \(u\equiv 0\) on \({\mathbb{R}}\cap S^+\). Define \(S^- = \left\{{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu {~\mathrel{\Big\vert}~}z\in S^+}\right\}\), and \begin{align*} U(z) \coloneqq \begin{cases} U(z) & z\in S^+ \\ -U(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) & z\in S^-. \end{cases} .\end{align*} Then \(U\) is a harmonic extension of \(u\) to \(S \coloneqq S^+ \cup(S^+ \cap{\mathbb{R}}) \cup S^-\). To see that \(U\) is harmonic on \(S\), it suffices to check that \(U\) satisfies the mean value property on \(S\). Clearly this holds in \(S^+\), so for \(z_0\in S^+\) we have \begin{align*} U(z_0) &= u(z_0) \\ &= {1\over 2\pi} \int_{-\pi}^\pi u(z_0 + re^{it} )\,dt\\ &= {1\over 2\pi} \int_{-\pi}^\pi U(z_0 + re^{it} )\,dt\\ .\end{align*} So for \(w_0\in S^-\), write it as \(w_0 = \mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu\), then \begin{align*} U(z_0) &\coloneqq-u(\mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu) \\ &= {1\over 2\pi }\int_{-\pi}^\pi - u\qty{\mkern 1.5mu\overline{\mkern-1.5muz_0 + re^{it} \mkern-1.5mu}\mkern 1.5mu} \,dt\\ &= {1\over 2\pi }\int_{-\pi}^\pi - u\qty{\mkern 1.5mu\overline{\mkern-1.5muz_0\mkern-1.5mu}\mkern 1.5mu + re^{-it} } \,dt\\ &= {1\over 2\pi }\int_{-\pi}^\pi - u\qty{\mkern 1.5mu\overline{\mkern-1.5muz_0 + re^{-it} \mkern-1.5mu}\mkern 1.5mu} \,dt\\ &= {1\over 2\pi }\int_{-\pi}^\pi U(z_0 + re^{it} ) \,dt .\end{align*}

Part 1: Let \(\Omega = \Omega^+ \cup I \cup\Omega^-\) be a region symmetric about \({\mathbb{R}}\). If \(f\) is holomorphic on \(\Omega^+\) extending continuously to \(I\) and real valued on \(I\), then \(f\) extends to a holomorphic function \(F\) on all of \(\Omega\) defined on \(\Omega^-\) by \(F(z) = \mkern 1.5mu\overline{\mkern-1.5muf(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu\).

Part 2: The map is \(T(z) = -i\qty{z+1\over z-1}\) with \(T^{-1}(z) = {z-i\over z+i}\), so \begin{align*} (T^{-1}\circ g \circ T)(z) &= T^{-1}\mkern 1.5mu\overline{\mkern-1.5mu\qty{-i {z+1\over z-1} }\mkern-1.5mu}\mkern 1.5mu \\ &= T^{-1}\qty{i{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + 1 \over \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - 1}} \\ &= {i\qty{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + 1 \over \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - 1} - i \over i\qty{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + 1 \over \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - 1} + i } \\ &= {(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + 1) - (\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - 1) \over (\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + 1) + (\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - 1)} \\ &= {1\over \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} .\end{align*}

Part 3: Define \(h: {\mathbb{H}}\to \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{H}}\mkern-1.5mu}\mkern 1.5mu\) by \(h(z) = (T\circ f\circ T^{-1})(z)\). Under \(T^{-1}: {\mathbb{D}}\to {\mathbb{H}}\), we have \(T(S^1) = {\mathbb{R}}\), so \(h\) is a holomorphic function on \({\mathbb{H}}\) that is continuous and real-valued on \({\mathbb{R}}\). By the reflection principle, defining \(H(z) \coloneqq\mkern 1.5mu\overline{\mkern-1.5muh(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu\) for \(\Im(z) < 0\) yields an entire function \(H: {\mathbb{C}}\to {\mathbb{C}}\) Noting that for \(g(z) \coloneqq\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\), \(g=g^{-1}\), we can write \begin{align*} H \coloneqq g^{-1}\circ h \circ = h^{-1}\circ (T^{-1}\circ f \circ T)\circ g .\end{align*} We can then conjugate \(H\) by \(T\) to get a direct formula in terms of \(f\), and unwinding this yields the extension \(F:{\mathbb{C}}\to {\mathbb{C}}\) defined by \begin{align*} F(z) = \begin{cases} f(z) & z\in {\mathbb{D}} \\ f_-(z) \coloneqq{1\over \mkern 1.5mu\overline{\mkern-1.5muf\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\mkern-1.5mu}\mkern 1.5mu} & z\in {\mathbb{D}}^c \\ f(z) = f_i(z) & z\in S^1 \end{cases} .\end{align*} In particular, \(H\) is an entire bounded function and thus constant, making \(F\) constant as well and consequently \(f\) is constant.

First, note that the Schwarz reflection principle can be applied here: let \(T: {\mathbb{D}}\to {\mathbb{H}}\) be the Cayley map, and consider \(\tilde f \coloneqq T\circ f \circ T^{-1}: {\mathbb{H}}\to {\mathbb{H}}\). Now \(T(S^1) = {\mathbb{R}}\), and since \(f(z)\in S^1\) when \(z\in S^1\), we have \(\tilde f({\mathbb{R}}) = {\mathbb{R}}\), i.e. this is a real-valued function on \({\mathbb{R}}\). So \(\tilde f\) extends holomorphically to \(\tilde F:{\mathbb{C}}\to CC\), and we can pull this back to a holomorphic extension of \(f\).

Extend \(f\) to \(F:{\mathbb{C}}\to {\mathbb{C}}\) by \(f(z) = 1/\mkern 1.5mu\overline{\mkern-1.5muf(1/\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\mkern-1.5mu}\mkern 1.5mu\) for \(z\in {\mathbb{D}}^c\), which generally has poles at the points \(1/\mkern 1.5mu\overline{\mkern-1.5muz_k\mkern-1.5mu}\mkern 1.5mu\) for \(z_k\in {\mathbb{D}}\) zeros of \(f\). Since \(f\) is nonvanishing, \(F\) has no poles and thus defines an entire function. By definition of \(F\), we have \(F({\mathbb{C}}) \subseteq f\qty{\left\{{{\left\lvert {z} \right\rvert} \leq 1}\right\}} \cup\mkern 1.5mu\overline{\mkern-1.5mu f\qty{\left\{{{\left\lvert {z} \right\rvert} \geq 1}\right\}}\mkern-1.5mu}\mkern 1.5mu\), which are both the continuous images of compact sets and thus compact and bounded. So \(F\) is a bounded entire function and thus constant.