# UMD Analysis Qualifying Exam/Jan08 Complex

## Problem 2

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

## Solution 2

Key steps

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

## Problem 4

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

## Solution 4

### Showing G 1:1 conformal mapping

First note that

{\displaystyle {\begin{aligned}(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{aligned}}}

Also, applying a , we have for all ${\displaystyle z_{1},z_{2}\in H\!\,}$ ,

${\displaystyle (2)\quad \sin z_{1}-\sin z_{2}=2\sin \left({\frac {z_{1}-z_{2}}{2}}\right)\cos \left({\frac {z_{1}+z_{2}}{2}}\right)\!\,}$

Hence if ${\displaystyle \sin z_{1}=\sin z_{2}\!\,}$ , then

${\displaystyle \sin \left({\frac {z_{1}-z_{2}}{2}}\right)=0\!\,}$

or

${\displaystyle \cos \left({\frac {z_{1}+z_{2}}{2}}\right)=0\!\,}$

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

${\displaystyle \sin \left({\frac {z_{1}-z_{2}}{2}}\right)=0\!\,}$

i.e.

${\displaystyle z_{1}=z_{2}\!\,}$

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

Therefore from ${\displaystyle (1)\!\,}$  and ${\displaystyle (2)\!\,}$ , ${\displaystyle g\!\,}$  is a ${\displaystyle 1:1\!\,}$  conformal mapping.

### Finding the domain D

To find ${\displaystyle D\!\,}$ , we only need to consider the image of the boundaries.

Consider the right hand boundary, ${\displaystyle C_{3}=\{z=x+iy|x={\frac {\pi }{2}},y>0\}\!\,}$

{\displaystyle {\begin{aligned}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{aligned}}}

Since ${\displaystyle y>0\!\,}$ ,

${\displaystyle g(C_{3})=(1,\infty )\!\,}$

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

{\displaystyle {\begin{aligned}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{aligned}}}

Since ${\displaystyle y>0\!\,}$ ,

${\displaystyle g(C_{2})=(-\infty ,-1)\!\,}$

Now consider the bottom boundary ${\displaystyle C_{1}=\{z=x+iy|-{\frac {\pi }{2}}<=x<={\frac {\pi }{2}},y=0\}\!\,}$ .

{\displaystyle {\begin{aligned}g(C_{1})&=g(x)\\&=\sin(x)\end{aligned}}}

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

${\displaystyle g(C_{2})=[-1,1]\!\,}$

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

{\displaystyle {\begin{aligned}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{aligned}}}

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

## Problem 6

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

## Solution 6

### Summation a_n Convergent

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

${\displaystyle \sum _{n=1}^{\infty }a_{n}=\infty \!\,}$

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

{\displaystyle {\begin{aligned}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{aligned}}}

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

${\displaystyle \Im (h(i))={\frac {1}{2}}\sum _{n=1}^{\infty }a_{n}\underbrace {(e^{n}-e^{-n})} _{E_{n}}\!\,}$

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

{\displaystyle {\begin{aligned}\Im (h(i))&\geq {\frac {1}{2}}(e^{1}-e^{-1})\underbrace {\sum _{n=1}^{\infty }a_{n}} _{=\infty }\\\\\\\Im (h(i))&\geq \infty \end{aligned}}}

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

### Show that h is analytic

In order to prove that ${\displaystyle h\!\,}$  is analytic, let us cite the following theorem

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

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

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