Fall 2019.6 #complex/qual/completed
A holomorphic mapping f:U→V is a local bijection on U if for every z∈U there exists an open disc D⊂U centered at z so that f:D→f(D) is a bijection. Prove that a holomorphic map f:U→V is a local bijection if and only if f′(z)≠0 for all z∈U.
- Inverse function theorem: if F∈C1(Rn→Rn) and Df is invertible at p, the F is invertible in a neighborhood of p, and F−1 is C1.
solution:
⟸: Let z∈U be fixed. Since f is holomorphic at z and f′(z)≠0, consider f(x,y) and its Jacobian as a real-valued function: Df=[uxuyvxvy]⟹det(Df)=uxvy−vxuy=u2x+v2x=|f′|2>0, so the derivative matrix is invertible at z. Applying the inverse function theorem yields that f is a smooth diffeomorphism on some neighborhood N∋p, and in particular is bijective on N.
⧸⟸: If f′(z)=0 for some z, then we claim that f can not be injective. Equivalently, injectivity of f implies f′≠0. Suppose f is holomorphic at z0 but f′(z0)=0. Write h(z):=f(z)−f(z0), which has a zero z0 of some order k≥2. For a disc D small enough about z0 avoiding the other (isolated) zeros of h and f′, for any p in a neighborhood of z0 and contained in D, ∫∂Df′(ξ)f(ξ)−pdξ=♯Z(f(z)−p), using the argument principle and that (f(ξ)−p)′=f′(ξ). But for D small enough, ♯Z(f(z)−p)=♯Z(f(z)−f(z0))=k by Rouché, so there are k solutions to f(z)=p. Since (f(z)−p)′≠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:=infz∈D|f(z)−w0|, 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.
solution:
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)?
solution (1):
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.
solution (2):
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.
solution (3):
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.
solution (4):
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.
solution (Cauchy-Riemann):
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*}
solution (Direct definition):
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})}.
solution (Power series):
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.
solution:
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.
solution:
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.
solution:
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.
solution:
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.