# Holomorphicity

## Fall 2019.6 #complex/qual/completed

A holomorphic mapping $$f: U \rightarrow V$$ is a local bijection on $$U$$ if for every $$z \in U$$ there exists an open disc $$D \subset U$$ centered at $$z$$ so that $$f: D \rightarrow f(D)$$ is a bijection. Prove that a holomorphic map $$f: U \rightarrow V$$ is a local bijection if and only if $$f^{\prime}(z) \neq 0$$ for all $$z \in U$$.

• Inverse function theorem: if $$F\in C^1({\mathbf{R}}^n\to {\mathbf{R}}^n)$$ and $$D_f$$ is invertible at $$p$$, the $$F$$ is invertible in a neighborhood of $$p$$, and $$F^{-1}$$ is $$C^1$$.

$$\impliedby$$: 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.

## Spring 2020 HW 1.7 #complex/qual/completed

Prove that $$f(z) = {\left\lvert {z} \right\rvert}^2$$ has a derivative at $$z=0$$ and nowhere else.

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 $$w\neq 0$$.

## Spring 2020 HW 1.8 #complex/qual/completed

Let $$f(z)$$ be analytic in a domain, and prove that $$f$$ is constant if it satisfies any of the following conditions:

• $${\left\lvert {f(z)} \right\rvert}$$ is constant.
• $$\Re(f(z))$$ is constant.
• $$\arg(f(z))$$ is constant.
• $$\overline{f(z)}$$ is analytic.

How do you generalize (a) and (b)?

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.

## Spring 2020 HW 1.9 #complex/qual/completed

Prove that if $$z\mapsto f(z)$$ is analytic, then $$z \mapsto \overline{f(\overline{z})}$$ is analytic.

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.

## Spring 2020 HW 1.10 #complex/qual/completed

• Show that in polar coordinates, the Cauchy-Riemann equations take the form \begin{align*} \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \text { and } \frac{\partial v}{\partial r}=-\frac{1}{r} \frac{\partial u}{\partial \theta} .\end{align*}

• Use (a) to show that the logarithm function, defined as \begin{align*} \operatorname{Log}z=\log r+i \theta \text { where } z=r e^{i \theta} \text { with }-\pi<\theta<\pi .\end{align*} is holomorphic on the region $$r> 0, -\pi < \theta < \pi$$.

Also show that this function is not continuous in $$r>0$$.

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.

## Fall 2021.1 #complex/qual/completed

Let $$f(z)$$ be an analytic function on $$|z|<1$$. Prove that $$f(z)$$ is necessarily a constant if $$f(\overline{z})$$ is also analytic.

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.

## Holomorphic functions form an integral domain #complex/qual/completed

Suppose $$D$$ is a domain and $$f, g$$ are analytic on $$D$$.

Prove that if $$fg = 0$$ on $$D$$, then either $$f \equiv 0$$ or $$g\equiv 0$$ on $$D$$.

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 $$D$$.

## Holomorphic functions with specified values #complex/qual/completed

Suppose $$f$$ is analytic on $${\mathbb{D}}^\circ$$. Determine with proof which of the following are possible:

• $$f\qty{1\over n} = (-1)^n$$ for each $$n>1$$.

• $$f\qty{1\over n} = e^{-n}$$ for each even integer $$n>1$$ while $$f\qty{1\over n} = 0$$ for each odd integer $$n>1$$.

• $$f\qty{1\over n^2} = {1\over n}$$ for each integer $$n>1$$.

• $$f\qty{1\over n} = {n-2 \over n-1}$$ for each integer $$n>1$$.

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 $$z_0 = 1$$.

#complex/qual/completed