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

{\bf Question}

\begin{itemize}
\item[a)]
Let $\ds f(z)=\sum_{n=-k}^\infty a_n(z-b)^n$ be the Laurent series
for a function analytic in the region $0<|z-b|<R$ and $k$ is a
non-negative integer or $\infty$.  In terms of this series explain
what is meant by

\begin{itemize}
\item[i)]
a removable singularity,

\item[ii)]
an isolated essential singularity,

\item[iii)]
a pole of order $m$.
\end{itemize}

Classify all the singularities of the function

$$\frac{(z^2-1)e^\frac{1}{z}}{(z^2-3z+2)\sin(z+1)}.$$


\item[b)]
Find the Laurent expansions of

$$\frac{1}{z^2(1-z)}$$

valid for

\begin{itemize}
\item[i)]
$0<|z|<1,$

\item[ii)]
$|z|>1,$

\item[iii)]
$0<|z-1|<1.$
\end{itemize}

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
Bookwork

$\ds f(z)=\frac{(z+1)(z-1)e^\frac{1}{z}}{(z-1)(z-2)\sin(z+1)}$

Removable singularity at $z=1$

Removable singularity at $z=-1$ because
$\ds\frac{z+1}{\sin(z+1)}\to1$ as $z\to-1$

Simple pole at $z=2$

Essential singularity at $z=0$ due to $e^\frac{1}{z}$.


\item[b)]

\begin{itemize}
\item[i)]
In $0<|z|<1$

$\ds\frac{1}{1-z}=1+z+z^2+\cdots$

so
$\ds\frac{1}{z^2(1-z)}=\frac{1}{z^2}+\frac{1}{z}+1+z+z^2+\cdots$


\item[ii)]
for $|z|>1$

$\ds\frac{1}{z^2}\frac{1}{1-z}=
\frac{1}{z^2}\frac{-1}{z\left(1-\frac{1}{z}\right)}=
\frac{-1}{z^3}\frac{1}{1-\frac{1}{z}}$

$\ds=\frac{-1}{z^3}\left(1+\frac{1}{z}+\frac{1}{z^2}+\cdots\right)=
-\frac{1}{z^3}-\frac{1}{z^4}-\frac{1}{z^6}-\cdots$


\item[iii)]
for $0<|z-1|<1$

$\ds\frac{1}{z^2}=\frac{1}{[1-(1-z)]^2}=
1+2(1-z)+3(1-z)^2+4(1-z)^3+\cdots$

so $\ds\frac{1}{z^2(1-z)}=\frac{-1}{z-1}+2-3(z-1)+4(z-1)^2-\cdots$

\end{itemize}

\end{itemize}

\end{document}