UMD Analysis Qualifying Exam/Jan08 Complex

Problem 2Edit

 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 2Edit

Key steps

• $\sin^2(\theta)=\frac{1-\cos(2\theta)}{2} \!\,$
• $\cos(z)=\sum_{n=1}^\infty \frac{(-1)^nz^{2n}}{(2n)!} \!\,$
• ratio test

Problem 4Edit

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

Solution 4Edit

Showing G 1:1 conformal mappingEdit

First note that

\begin{align} (1) \quad |g^{\prime}(z)| &= |\sin^{\prime}(z)|\\ &= |\cos(z)| \\ &= \left|\frac{e^{-iz}+e^{iz}}{2}\right| \\ &\geq \frac{1}{2} (|e^{-ix}||e^{y}|-|e^{ix}||e^{-y}| ) \\ &\geq \frac{1}{2}(e^y-e^{-y}) \\ &> 0 \quad \mbox{since }y>0 \end{align}

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

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

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 DEdit

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{align} g(C_3) &= g\left(\frac{\pi}{2}+ yi\right)\\ &= \sin\left(\frac{\pi}{2}+ yi\right) \\ &= \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{align}

Since $y>0 \!\,$,

$g(C_3)=(1,\infty) \!\,$

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

\begin{align} g(C_2) &= g\left(-\frac{\pi}{2}+ yi\right)\\ &= \sin\left(-\frac{\pi}{2}+ yi\right) \\ &= \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{align}

Since $y>0 \!\,$,

$g(C_2)=(-\infty,-1) \!\,$

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

\begin{align} g(C_1) &= g(x)\\ &= \sin(x) \end{align}

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{align} 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 \mbox{Upper Half Plane} \end{align}

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

Problem 6Edit

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

Solution 6Edit

Summation a_n ConvergentEdit

We want to show that $\sum_{n=1}^\infty a_n \!\,$ is convergent. Assume for the sake of contradiction that $\sum_{n=1}^\infty a_n \!\,$ is divergent i.e.

$\sum_{n=1}^\infty a_n = \infty\!\,$

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

\begin{align} h(i) &= \sum_{n=1}^\infty a_n \sin(ni) \\ &= \sum_{n=1}^\infty a_n \frac{e^{-n}-e^n}{2i} \\ &= \sum_{n=1}^\infty a_n \frac{-i(e^{-n}-e^n)}{2} \\ &= \sum_{n=1}^\infty a_n \frac{i(e^n-e^{-n})}{2} \end{align}

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

$\Im(h(i)) = \frac12 \sum_{n=1}^\infty a_n \underbrace{(e^n-e^{-n})}_{E_n} \!\,$

The sequence $\{E_n\} \!\,$ is increasing ($E_1 < E_2 < \ldots < E_n < E_{n+1} \!\,$) 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{align} \Im(h(i)) &\geq \frac12 (e^1 - e^{-1}) \underbrace{\sum_{n=1}^\infty a_n}_{= \infty} \\ \\ \\ \Im(h(i)) &\geq \infty \end{align}

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

Show that h is analyticEdit

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 $h_n(z)= \sum_{k=1}^{n} a_k \sin(k z)\!\,$.   Let $K\!\,$ be a compact set of $U=\{\Im{z}>0\}\!\,$.   Since $\sin(kz) \!\,$ is continuous