UMD Analysis Qualifying Exam/Jan08 Complex

Solution 2
Key steps


 * $$\sin^2(\theta)=\frac{1-\cos(2\theta)}{2} \!\,$$


 * $$\cos(z)=\sum_{n=1}^\infty \frac{(-1)^nz^{2n}}{(2n)!} \!\,$$


 * ratio test

Showing G 1:1 conformal mapping
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 D
To find $$ D \!\,$$, we only need to consider the image of the boundaries.

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

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

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

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