\documentclass[12pt]{article}
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\begin{document}

{\bf Question}

\begin{itemize}
\item[a)]
Evaluate the sum of the series

$$\sum_{n=-\infty}^\infty\frac{1}{(2n-1)^2}$$

You should justify the steps in your method, except that you may
assume without proof inequalities relating to the function
$\cot\pi z$.

\item[b)]
State Rouche's theorem and use it to show that all the roots of
the equation

$$z^7+(1-i)z^5+2z^3-1=0$$

lie in the annulus $\frac{1}{2}\leq|z|<2$.  Find a value of $k$
smaller than 2 with the property that all the roots of the
equation satisfy $|z|<k$.

\end{itemize}



\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$\ds S=\sum_{-\infty}^{\infty}\frac{1}{(2n-1)^2}=
\frac{1}{4}\sum_{n=-\infty}^{\infty}\frac{1}
{\left(n-\frac{1}{2}\right)^2}$

The function $\ds f(z)=\frac{\pi\cot\pi
z}{\left(z-\frac{1}{2}\right)^2}$ has a simple pole at $z=n\in{\bf
N}$ with residue $\ds\frac{1}{\left(n-\frac{1}{2}\right)^2}$,


since $\ds(z-n)f(z)=
\frac{\pi(z-n)}{\sin\pi(z-n)}\frac{\cos\pi(z-n)}
{\left(z-\frac{1}{2}\right)^2}\rightarrow
\frac{1}{\left(n-\frac{1}{2}\right)^2}$ as $z\to n$.

$f(z)$ has a pole of order 2 at $z=\frac{1}{2}$.  The residue is
given by

$\ds\lim_{z\to\frac{1}{2}}\frac{d}{dz}
\left(\left(z-\frac{1}{2}\right)^2f(z)\right)=
\lim_{z\to\frac{1}{2}}\frac{d}{dz}\pi\cot\pi z$

$\ds\hspace{0.2in}=-\pi^2\csc^2\frac{1}{2}\pi=-\pi^2$

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

On $C_N \,\,\, \pi\cot\pi z$ is uniformly bounded (by $K$).

So $\ds\left|\int_{C_N}f(z)dz\right|\leq
\frac{K8\left(N+\frac{1}{2}\right)}{N^2}\rightarrow0$ as
$N\to\infty$.

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

so letting $\ds N\to\infty, \,\,\,
\sum_{n=-\infty}^\infty\frac{1}{\left(n-\frac{1}{2}\right)^2}=
\pi^2$.  Thus $S=\frac{\pi^2}{4}$.

\item[b)]
Rouche's Theorem states that 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 $C$

\begin{itemize}
\item[i)]
Let $f(z)=-1, \,\,\, g(z)=z^7+(1-i)z^5+2z^3$,

for $|z|=\frac{1}{2} \,\,\,
|g(z)|\leq(\frac{1}{2})^7+\sqrt2(\frac{1}{2})^5+2(\frac{1}{2})^3<1=|f(z)|$

$f(z)$ has no zeros inside $|z|=\frac{1}{2}$, so $f(z)+g(z)$ has
none inside $|z|=\frac{1}{2}$.

\item[ii)]
Let $f(z)=z^7, \,\,\, g(z)=(1-i)z^5+2z^3-1$,

for $|z|=2 \,\,\,
|g(z)|\leq\sqrt22^5+2^4+1<2^6+2^4+1=81<2^7=|f(z)|$

$f(z)$ has 7 zeros inside $|z|=2$, and so does $f(z)+g(z)$
therefore.

For $|z|=1.6 \,\,\, |g(z)|<\sqrt2(1.6)^5+2(1.6)^4+1\approx24.02$

$\hspace{1in}|f(z)|=(1.6)^7\approx26.84$  So $a=1.6$ will do.

(This doesn't quite work with $a=1.5$).

\end{itemize}

\end{itemize}


\end{document}