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

{\bf Question}

\begin{itemize}
\item[a)]
Find the Taylor series expansion of the function
$f(z)=(z+1)/(z-1)$ about the origin.

Find the Laurent expansion of $f(z)$ about the origin for $|z|>1$.

Find the Laurent expansion of $f(z)$ about $z=1$.

\item[b)]
Locate the zeros and singularities of the function

$$\frac{z^2(z^2-z+1)\exp(1/z)}{z^3-13z^2+5z+7}$$

Classify the singularities, and determine the behaviour of the
function at infinity.
\end{itemize}



\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
For $\ds|z|<1, \,\,\, \frac{1}{1-z}=1+z+z^2+\cdots$

So
$\ds\frac{z+1}{z-1}=-(1+z)(1+z+z^2+\cdots)=-(1+2z+2z^2+2z^3+\cdots)$,
this is the required Taylor expansion

For $\ds|z|>1 \,\,\,
\frac{z+1}{z-1}=\frac{z+1}{z\left(1-\frac{1}{z}\right)}=
\left(1+\frac{1}{z}\right)
\left(1+\frac{1}{z}+\frac{1}{z^2}+\cdots\right)$

$\ds\hspace{0.5in}=
1+\frac{2}{z}+\frac{2}{z^2}+\frac{2}{z^3}+\cdots$ this is the
required Laurent expansion about the origin.

Now $\ds\frac{z+1}{z-1}=\frac{z-1+2}{z-1}=1+\frac{2}{z-1}$, this
is the expansion about $z=1$, having only two terms.

\item[b)]
$z^2-z+1=0$ when $z=\frac{1\pm i\sqrt3}{2}$.  These are zeros of
the function.

$z^3-13z^2+5z+7=(z-1)(z^2-12z-7)=0$ when $z=1$ and
$z=6\pm\sqrt{43}$.

So the function has simple poles at $z=1$ and $z=6\pm\sqrt{43}$.

The function has an essential singularity at $z=0$.

To investigate the behaviour at infinity, replace $z$ by
$\frac{1}{z}$ to obtain.

$\ds\frac{\frac{1}{z^2}\left(\frac{1}{z^2}-\frac{1}{z}+1\right)\exp(z)}
{\frac{1}{z^3}-\frac{13}{z^2}+\frac{5}{z}+7}=
\frac{(1-z+z^2)e^z}{z(1-13z+5z^2+7z^3)}$

This has a simple pole at $z=0$, so the original function has a
simple pole at infinity.

\end{itemize}

\end{document}
