Holomorphicity

\(\impliedby\): 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 \({ \overline{{\partial}}}_z f = 0\), but \begin{align*} { \overline{{\partial}}}_z {\left\lvert {z} \right\rvert}^2 = { \overline{{\partial}}}_z z\overline{z} = 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\overline{z} \over z} = \overline{z} \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} \\ &= \overline{w}(t+1) \\ &\overset{t\to 1}\to 2\overline{w} ,\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 \(\overline{f} = u +i\tilde v\) 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 \(g(z) \coloneqq\overline{f(\overline{z})}\) 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 \(g(z) \coloneqq(f(z^*))^* \coloneqq\overline{f(\overline{z})}\), 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 \(g'(z) \coloneqq\overline{f'(\overline{z})}\).

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 \overline{z^k} }^* = \sum_{k\geq 0} \overline{c_k} 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 \(\tilde f(z) \coloneqq f(\overline{z})\). 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 \(z_0 = 1\).