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 $$\begin{align*} {\left\lvert {x-y} \right\rvert}<\delta \implies {\left\lvert {f(x) - f(y)} \right\rvert} < {\varepsilon}&& \forall x, y\in {\mathbf{R}} .\end{align*}$$ 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 ${\mathbf{C}}$, 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 $$\begin{align*} d: A \times B &\to {\mathbf{R}}\\ (x, y) &\mapsto {\left\lVert {x- y} \right\rVert} .\end{align*}$$ 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: $$\begin{align*} S \coloneqq\left\{{t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}E \cap(-\infty, t), E \cap(t, \infty) \text{ are uncountable}}\right\} \subseteq {\mathbf{R}} .\end{align*}$$ Write $$\begin{align*} S_- &\coloneqq\left\{{ t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}E \cap(- \infty, t) \text{ is countable}}\right\} \\ S_+ &\coloneqq\left\{{ s\in {\mathbf{R}}{~\mathrel{\Big\vert}~}E \cap(s, \infty) \text{ is countable}}\right\} .\end{align*}$$

Note that $S_- \neq {\mathbf{R}}$ since then we could write $E = \bigcup_{n\in {\mathbf{Z}}} E \cap(- \infty, n)$ 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 $\gamma \subseteq {\mathbf{C}}$ is open in ${\mathbf{C}}$.

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 $\gamma \coloneqq\left\{{t e^{i\theta_0}{~\mathrel{\Big\vert}~}t\in {\mathbf{R}}}\right\}$ which has no open subsets.

Note that this implies that any ${\mathbf{R}}{\hbox{-}}$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 $f: {\mathbf{R}}^2\to {\mathbf{R}}$ 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 ${\mathbf{C}}$ 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 $U \coloneqq{\mathbf{C}}\setminus(-\infty, 0]$ 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*}

\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*}

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 $k\in {\mathbf{Z}}$, and this is a pole of order 2 for $f$. Every $k\in {\mathbf{Z}}$ 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 $g$, $$\begin{align*} g(z+1) = \sum_{k\in {\mathbf{Z}}} ((z-1)-k)^{-2} = \sum_{k'\in {\mathbf{Z}}} (z-k)^{-2} ,\end{align*}$$ where $k' \coloneqq k+1$, and the equality is true since both sums run over all of ${\mathbf{Z}}$.

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 ${\mathbf{C}}$. 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

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

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

• Let $I$ be the integral over ${\mathbf{R}}$. 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 ${\mathbf{R}}$ and $\gamma_2$ is the usual half-circle of radius $R$.

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

  • 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 ${\mathbf{C}}$ 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 $T \subseteq {\mathbf{C}}\setminus\gamma$ 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 ${\mathbf{C}}$ 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 $x\in {\mathbf{R}}$, show $f$ is holomorphic in $D$ by Morera’s theorem.
• Let $\Delta \subset D$ be a triangle in $D$.
• Case 1: If $\Delta \cap{\mathbf{R}}= 0$, then $f$ is holomorphic on $\Delta$ and $\int_\Delta f = 0$.
• Case 2: one side or vertex of $\Delta$ intersects ${\mathbf{R}}$, 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 $\Delta_{\varepsilon}\cap{\mathbf{R}}= 0$ 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 ${\mathbf{C}}$, 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 $D_{\varepsilon}(z_0)\to {\mathbf{R}}$. 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 ${\mathbb{D}}_r \subseteq \bigcup_{k=0}^{n-1} \zeta_n^k S_r$, 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 ${\left\lvert {f(z)} \right\rvert} \in [0, M] \subseteq {\mathbf{R}}$ 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 ${\mathbf{C}}$, 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 ${\mathbf{C}}$. 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 $z\in {\mathbf{C}}$, 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 ${\mathbf{C}}$.

  • 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 ${\mathbf{C}}$ and thus $1/p$ is bounded above on ${\mathbf{C}}$.

• 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 ${\mathbf{C}}\setminus Z_f$.
• 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 ${\mathbf{C}}$
  • If not, then ${\left\lvert {\tilde g(0)} \right\rvert} = 1+{\varepsilon}> 1$, but then there would be a domain $\Omega \subseteq {\mathbf{C}}\setminus\left\{{0}\right\}$ 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 $c\in {\mathbf{C}}$ 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 $g(U) \subseteq {\mathbf{C}}$ 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 $g(U) \subseteq ({\mathbf{C}}\setminus D_{A\over R^n}(0))$ 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 ${\mathbf{C}}$, 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 $Z_n \coloneqq\left\{{z\in {\mathbf{C}}{~\mathrel{\Big\vert}~}f^{(n)}(z) = 0 }\right\}$, then by hypothesis $\bigcup_{n\geq 0} Z_n = {\mathbf{C}}$. 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 $\gamma_1 \coloneqq\left\{{it {~\mathrel{\Big\vert}~}t\in {\mathbf{R}}}\right\}$ 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 $z=1 \in {\mathbf{R}}$ 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 ${\mathbf{C}}$), 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 $\left\{{1+it{~\mathrel{\Big\vert}~}t\in {\mathbf{R}}}\right\}$, 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 $k\in {\mathbf{Z}}$. 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 ${\mathbf{R}}$ or $i{\mathbf{R}}$. 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 $f({\mathbf{R}}_{\geq 0}) \subseteq {\mathbf{R}}_{\geq 0}$.
• $\Delta_{\gamma_2}\operatorname{Arg}(f) = 4\cdot {\pi\over 2} = 2\pi$ 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: $$\begin{align*} \oint_T g(z)\,dz &\coloneqq\oint_T \int_{\mathbf{R}}f(t)e^{-izt} \,dt\,dz\\ &\coloneqq\int_{\mathbf{R}}\oint_T f(t)e^{-izt} \,dz\,dt\\ &\coloneqq\int_{\mathbf{R}}f(t) \qty{ \oint_T e^{-izt} \,dz} \,dt\\ &\coloneqq\int_{\mathbf{R}}f(t) \cdot 0 \,dt\\ &= 0 ,\end{align*}$$ 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 $i{\mathbf{R}}$ 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 $i{\mathbf{R}}$ and the right of $S$. The map preserves $i{\mathbf{R}}$ and as $t$ traces out $(-\infty, 0^-, 0^+, \infty)$, $f(it)$ traces out $(0^+, \infty, -\infty, 0^-)$, so this preserves the orientation of $i{\mathbf{R}}$. 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 $a\in {\mathbf{R}}$, $\psi_a({\mathbf{R}}) = {\mathbf{R}}$ 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 $i{\mathbf{R}}$ 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 ${\mathbf{R}}$, so the image regions are symmetric about ${\mathbf{R}}$, and it suffices to find the angle of the line $f(C_1)$. Note that $C_1 \cap i{\mathbf{R}}= ic$, 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 $f({\mathbf{R}}) = {\mathbf{R}}$ and $f(i{\mathbf{R}}) = i {\mathbf{R}}$. 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 $f(i{\mathbf{R}})$: 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 $f(i{\mathbf{R}})$ is oriented from bottom to top, as in the image.

That the region outside the disc is mapped to the strip shown: points $x\in {\mathbf{R}}$ 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 $i{\mathbf{R}}$, so the new region should be on the right with respect to both $f(i{\mathbf{R}})$ 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 ${\mathbf{C}}\setminus[0, 1]$.
• Move the slit with $z\mapsto 4(z-1/2)$ to get ${\mathbf{C}}\setminus[-2, 2]$.
• Apply $z\mapsto z+z^{-1}$ to get ${\mathbf{C}}\setminus{\mathbb{D}}$, 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 $f: {\mathbf{C}}\setminus\left\{{0}\right\}\to {\mathbb{D}}\setminus\left\{{0}\right\}$ 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 $\pm 1\in {\mathbf{R}}$. 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 $f({\mathbf{C}}\setminus{\mathbb{D}}) = {\mathbf{C}}\setminus[-1, 1]$.

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 $\Omega \coloneqq{\mathbf{C}}\setminus({\mathbb{D}}\cup\left\{{\pm 1}\right\})$ 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 ${\mathbf{C}}\setminus[-1,1]\to {\mathbb{D}}$.

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

• 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 $G:\Delta\to {\mathbf{C}}$, 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 $T:{\mathbf{C}}\to {\mathbb{D}}$ for the Cayley map, then $F\coloneqq f\circ T$ satisfies $F({\mathbf{C}}) = T(f({\mathbf{C}})) \subseteq T({\mathbb{H}}) = {\mathbb{D}}$, 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 $$\begin{align*} F: {\mathbb{D}}&\to {\mathbf{C}}\\ z &\mapsto {f(Rz) \over M} .\end{align*}$$ 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: $$\begin{align*} {\left\lvert {(f^{-1}\circ g)(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert} \implies {\left\lvert {g(z)} \right\rvert} \leq {\left\lvert {f(z)} \right\rvert} \qquad \forall {\mathbb{D}}\in {\mathbf{Z}} .\end{align*}$$ 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 $z, f(z)\in {\mathbf{R}}$, $F$ 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 $F:{\mathbf{CP}}^1\to {\mathbf{CP}}^1$ 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 ${\mathbf{C}}$ 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 $f: U\to {\mathbf{C}}$. 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 ${\mathbf{Z}}{\hbox{-}}$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 ${\mathbf{Z}}{\hbox{-}}$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 $w$.