\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}

{\bf Question}

\begin{itemize}
\item[a)]
Prove that

$$\sum_{n=-\infty}^\infty\frac{1}{(a-n)^2}=\frac{\pi^2}{\sin^2\pi
a},$$

where $a$ is not an integer.


\item[b)]
State Rouche's Theorem.

Suppose that $g(z)$ is analytic inside and on the unit circle
$|z|=1$, and that $|g(z)|<1$ for $z$ on this circle.  Show that
there is a unique point $z_0$ inside the circle for which
$g(z_0)=z_0$.

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
The function $\ds f(z)=\frac{\pi\cot\pi z}{(a-z)^2}$ has simple
poles at $z=0,\pm1\cdots$.

At $z=n$ the residue is $\ds\frac{1}{(a-n)^2}$.

The function has a pole of order 2 at $z=a$, and we have to
calculate the residue.

By Taylor's theorem

$\pi\cot\pi z=\pi\cot\pi a-\pi^2\csc^2\pi a(z-a)+$ terms of higher
degree in $(z-a)$.

So $\ds\frac{\pi\cot\pi z}{(z-a)^2}= \frac{\pi\cot\pi
a}{(z-a)^2}-\frac{\pi^2\csc^2\pi a}{(z-a)}+g(z)$ (which is
analytic)

So the residue is $-\pi^2\csc^2\pi a$

${}$

or, we can use the formula
res$\ds(f,a)=\left.\frac{d}{dz}(z-a)^2f(z)\right|_{z=a}$

${}$

Now let $C_N$ be the square with vertices $\pm(N+\frac{1}{2})(1\pm
i) \hspace{0.2in} N\geq0$

On the upper sides parallel to the real axis
$z=x+(N+\frac{1}{2})i$.

$\ds|\cot\pi z|=\left|\frac{\cos\pi z}{\sin\pi z}\right|=
\left|i\frac{e^{i\pi z}+e^{-i\pi z}}{e^{i\pi z}-e^{-i\pi
z}}\right|=\left|\frac{e^{2i\pi z}+1}{e^{2i\pi z}-1}\right|$

$\ds\leq \frac{1+|e^{2\pi iz|}}{1-|e^{2\pi iz|}}=
\frac{1+e^{-2\pi(N+\frac{1}{2})}}{1-e^{-2\pi(N+\frac{1}{2})}}\leq
\frac{2}{1-e^{-\pi}} (N\geq0)$ for all $N$.

Also since $|\cot\pi z|=|\cot\pi(-z)|$ the same bound serves on
the bottom side of the square.

On the sides parallel to the imaginary axis
$z=\pm(N+\frac{1}{2})+iy$.

$|\cot\pi z|=|\cot\pi(\pm N+\frac{1}{2}+iy)|$

$=|\cot\pi(\frac{1}{2}+iy)|=|-\tan\pi iy|=|\tanh y|\leq 1$

So $\exists K$ independent of $N$, such that $|\pi\cot\pi z|\leq
K$ for $z\in C_N$.

Now provided $N\geq|a|$, we have

$\ds\int_{C_N}f(z)dz=2\pi i\left\{\sum_{n=-N}^N\frac{1}{(a-n)^2}-
\pi^2\csc^2\pi a\right\}$

Now $\ds\left|\int_{C_N}\frac{\pi\cot\pi z}{(a-z)^2}dz\right|\leq
\frac{K8\left(N+\frac{1}{2}\right)}
{\left(N+\frac{1}{2}-|a|\right)^2}\rightarrow 0$ as $N\to\infty$.

since $|z|\geq N+\frac{1}{2}$ on $C_N$.

Letting $N\to\infty$ therefore gives
$\ds\sum_{n=-\infty}^\infty\frac{1}{(a-n)^2}=
\frac{\pi^2}{\sin^2\pi a}$


\item[b)]
Rouche's Theorem is as follows:

If $f(z)$ and $g(z)$ are both analytic inside and on the closed
contour $C$, and if $|g(z)|<|f(z)|$ on $C$ then $f(z)$ and
$F(z)+g(z)$ have the same number of zeros inside $C$, (counting
multiplicities).

Let $f(z)=-z$.  Then for $z$ on $C$

$|f(z)|=|z|=1>|g(z)|$

SO by Rouche's theorem $f(z)$ and $f(z)+g(z)$ have the same number
of zeros inside $C$.  $f(z)=0$ only for $z=0$ - a simple zero, so
$g(z)-z=0$ has a unique solution $z_0$ inside $C$.

i.e. $g(z_0)=z_0$.

\end{itemize}

\end{document}
