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

{\bf Question}

\begin{itemize}
\item[a)]
Show that if $f(z)$ has a pole of order 2 at $z=a$, then its
residue at $z=a$ is

$$\lim_{z\to a}\left(\frac{d}{dz}((z-a)^2f(z))\right).$$


\item[b)]
The function $f(z)$ is regular except for $z=0$ and $z=i$.  Find
$f(z)$ when

\begin{itemize}
\item[i)]
$f(z)$ has a simple pole at $z=0$,

\item[ii)]
$f(z)$ has a pole of order 2 with residue 5 at $z=i$,

\item[iii)]
$\ds\lim_{z\to\infty}zf(z)=3$ and

\item[iv)]
$f(-i)=0$.

\end{itemize}

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
Bookwork from the Laurent Series


\item[b)]
We look for a function of the form

$\ds f(z)=\frac{P(z)}{z(z-i)^2}$ using (i) and (ii).

since $\ds\lim_{z\to\infty}zf(z)$ is finite and non zero, $P$ is
quadratic.

$\ds\lim_{z\to\infty}\frac{z(\alpha z^2+\beta
z+\gamma)}{z(z-i)^2}=3$, therefore $\alpha=3$.

The residue at $z=i$ is

$\ds\lim_{z\to i}\frac{d}{dz}\frac{3z^2+\beta
z+\gamma}{z}=\lim_{z\to i}3-\frac{\gamma}{z^2}=3+\gamma=5$ so
$\gamma=2$

$f(-i)=0$ so $3(-i)^2+\beta(-i)+2=0$

$\beta(-i)-1=0$ so $\beta=i$

Thus $\ds f(z)\frac{3z^2+iz+2}{z(z-1)^2}$

\end{itemize}

\end{document}
