UMD Analysis Qualifying Exam/Jan08 Complex

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

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 trigonometric identity, 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.

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

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 (${\displaystyle E_{1}<E_{2}<\ldots <E_{n}<E_{n+1}\!\,}$) 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.

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