UMD Analysis Qualifying Exam/Jan08 Complex

Problem 2

Prove there is an entire function $f$ so that for any branch $g$ of $\sqrt{z}$ ${\sin }^2(g(z))=f(z)$ for all $z$ in the domain of definition of $g$

Solution 2

Key steps

• ${\sin }^2(\theta )=\frac{1-\cos (2\theta )}{2}$

• $\cos (z)=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n{z}^{2n}}{(2n)!}$

• ratio test

Problem 4

Let $H$ be the domain $\{z:-\frac{\pi }{2}<\mathrm{\Re }(z)<\frac{\pi }{2},\mathrm{\Im }(z)>0\}$. Prove that $g=\sin (z)$ is a 1:1 conformal mapping of $H$ onto a domain $D$. What is $D$?

Solution 4

Showing G 1:1 conformal mapping

First note that

$\begin{array}{cc}(1)|g^{\prime }(z)|& =|{\sin }^{\prime }(z)|\\ & =|\cos (z)|\\ & =\left|\frac{e^{-iz}+e^{iz}}{2}\right|\\ & \ge \frac{1}{2}(|e^{-ix}||e^y|-|e^{ix}||e^{-y}|)\\ & \ge \frac{1}{2}(e^y-e^{-y})\\ & >0\text{since }y>0\end{array}$

Also, applying a trigonometric identity, we have for all $z_1,z_2\in H$,

$(2)\sin z_1-\sin z_2=2\sin \left(\frac{z_1-z_2}{2}\right)\cos \left(\frac{z_1+z_2}{2}\right)$

Hence if $\sin z_1=\sin z_2$, then

$\sin \left(\frac{z_1-z_2}{2}\right)=0$

or

$\cos \left(\frac{z_1+z_2}{2}\right)=0$

The latter cannot happen in $H$ since $|g^{\prime }(z)|=|\cos (z)|>0$ so

$\sin \left(\frac{z_1-z_2}{2}\right)=0$

i.e.

$z_1=z_2$

Note that the zeros of $\sin (z)=\sin (x+iy)$ occur at $x=k\pi ,k\in \mathbb{Z}$. Similary the zeros of $\cos (z)=\cos (x+iy)$ occur at $x=\frac{\pi }{2}+k\pi ,k\in \mathbb{Z}$.

Therefore from $(1)$ and $(2)$, $g$ is a $1:1$ conformal mapping.

Finding the domain D

To find $D$, we only need to consider the image of the boundaries.

Consider the right hand boundary, $C_3=\{z=x+iy|x=\frac{\pi }{2},y>0\}$

$\begin{array}{cc}g(C_3)& =g(\frac{\pi }{2}+yi)\\ & =\sin (\frac{\pi }{2}+yi)\\ & =\frac{e^{i(\frac{\pi }{2}+yi)}-e^{-i(\frac{\pi }{2}+yi})}{2i}\\ & =\frac{e^{i\frac{\pi }{2}}{e}^{-y}-e^{-i\frac{\pi }{2}}{e}^y}{2i}\\ & =\frac{e^{i\frac{\pi }{2}}{e}^{-y}+e^{i\frac{\pi }{2}}{e}^y}{2i}\\ & =\frac{e^{i\frac{\pi }{2}}(e^{-y}+e^y)}{2i}\\ & =\frac{i(e^{-y}+e^y)}{2i}\\ & =\frac{e^{-y}+e^y}{2}\\ \end{array}$

Since $y>0$,

$g(C_3)=(1,\mathrm{\infty })$

Now, consider the left hand boundary $C_2=\{z=x+iy|x=-\frac{\pi }{2},y>0\}$.

$\begin{array}{cc}g(C_2)& =g(-\frac{\pi }{2}+yi)\\ & =\sin (-\frac{\pi }{2}+yi)\\ & =\frac{e^{i(-\frac{\pi }{2}+yi)}-e^{-i(-\frac{\pi }{2}+yi})}{2i}\\ & =\frac{e^{-i\frac{\pi }{2}}{e}^{-y}-e^{i\frac{\pi }{2}}{e}^y}{2i}\\ & =\frac{-e^{i\frac{\pi }{2}}{e}^{-y}-e^{i\frac{\pi }{2}}{e}^y}{2i}\\ & =\frac{e^{i\frac{\pi }{2}}(-e^{-y}-e^y)}{2i}\\ & =-\frac{i(e^{-y}+e^y)}{2i}\\ & =-\frac{e^{-y}+e^y}{2}\\ \end{array}$

Since $y>0$,

$g(C_2)=(-\mathrm{\infty },-1)$

Now consider the bottom boundary $C_1=\{z=x+iy|-\frac{\pi }{2}<=x<=\frac{\pi }{2},y=0\}$.

$\begin{array}{cc}g(C_1)& =g(x)\\ & =\sin (x)\end{array}$

Since $-\frac{\pi }{2}<=x<=\frac{\pi }{2}$,

$g(C_2)=[-1,1]$

Hence, the boundary of $H$ maps to the real line. Using the test point $z=i$, we find

$\begin{array}{cc}g(i)& =\sin (i)\\ & =\frac{e^{i(i)}-e^{-i(i)}}{2i}\\ & =\frac{e^{-1}-e^1}{2i}\\ & =\frac{-i(e^{-1}-e^1)}{2}\\ & =\frac{i(e^1-e^{-1})}{2}\\ & =\frac{i}{2}(e-\frac{1}{e})\\ & \in \text{Upper Half Plane}\end{array}$

We then conclude $D=g(H)=\text{Upper Half Plane}$

Problem 6

Suppose that for a sequence $a_n\in R$ and any $z,\mathrm{\Im }(z)>0$, the series $h(z)=\sum _{n=1}^{\mathrm{\infty }}{a}_n\sin (nz)$ is convergent. Show that $h$ is analytic on $\{\mathrm{\Im }(z)>0\}$ and has analytic continuation to $C$

Solution 6

Summation a_n Convergent

We want to show that $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is convergent. Assume for the sake of contradiction that $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ is divergent i.e.

$\sum _{n=1}^{\mathrm{\infty }}{a}_n=\mathrm{\infty }$

Since $h(z)$ is convergent in the upper half plane, choose $z=i$ as a testing point.

$\begin{array}{cc}h(i)& =\sum _{n=1}^{\mathrm{\infty }}{a}_n\sin (ni)\\ & =\sum _{n=1}^{\mathrm{\infty }}{a}_n\frac{e^{-n}-e^n}{2i}\\ & =\sum _{n=1}^{\mathrm{\infty }}{a}_n\frac{-i(e^{-n}-e^n)}{2}\\ & =\sum _{n=1}^{\mathrm{\infty }}{a}_n\frac{i(e^n-e^{-n})}{2}\end{array}$

Since $h(i)$ converges in the upper half plane, so does its imaginary part and real part.

$\mathrm{\Im }(h(i))=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}{a}_n\underset{E_n}{\underbrace{(e^n-e^{-n})}}$

The sequence $\{E_n\}$ is increasing ($E_1<E_2<\dots <E_n<E_{n+1}$) since $e^{n+1}>e^n$ and $e^{-n}>e^{-(n+1)}$ e.g. the gap between $e^n$ and $e^{-n}$ is grows as $n$ grows. Hence,

$\begin{array}{cc}\mathrm{\Im }(h(i))& \ge \frac{1}{2}(e^1-e^{-1})\underset{=\mathrm{\infty }}{\underbrace{\sum _{n=1}^{\mathrm{\infty }}{a}_n}}\\ \\ \\ \mathrm{\Im }(h(i))& \ge \mathrm{\infty }\end{array}$

This contradicts that $h(i)$ is convergent on the upper half plane.

Show that h is analytic

In order to prove that $h$ is analytic, let us cite the following theorem

Theorem Let $h_n$ be a sequence of holomorphic functions on an open set $U$.   Assume that for each compact subset $K$ of $U$ the sequence converges uniformly on $K$, and let the limit function be $h$.   Then $h$ is holomorphic.

Proof See Theorem 1.1 in Chapter V, Complex Analysis Fourth Edition, Serge Lang.

Now, define $h_n(z)=\sum _{k=1}^n{a}_k\sin (kz)$.   Let $K$ be a compact set of $U=\{\mathrm{\Im }z>0\}$.   Since $\sin (kz)$ is continuous