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

{\bf Question}

\begin{itemize}
\item[i)]
Let $\ds h(z)=\frac{1}{(z-2)(z+3)}$.

Find the Laurent series expansion of $f(z)$ in powers of $z$,
valid in the annulus $2<|z|<3$ and obtain the coefficient of $z^n$
explicitly for $n=-2,-1,0,1,2.$


\item[ii)]
Let $f(z)$ and $g(z)$ be analytic functions in a neighbourhood of
$z=a$.  Let $f(z)$ have a zero of order $k$ at $z=a$ and $g(z)$
have a zero of order $l$ at $z=a$.  If $k>l$ show that
$F(z)=\frac{f(z)}{g(z)} \hspace{0.2in} (z\not=a)$ has a removable
singularity at $z=a$ and that if we extend its definition to $z=a$
by defining $F(a)=0$ then $F$ has a zero of order $k-l$ at $z=a$.

Describe the nature of $F(z)$ near $z=a$ if $k<l$.

If $k=l$ show that $F(z)$ again has a removable singularity at
$z=a$.  If $g'(a)\not=0$ prove that $l=1$ and

$$\lim_{z\to a}\frac{f(z)}{g(z)}=\frac{f'(a)}{g'(a)}.$$

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[i)]
$\ds\frac{1}{(z-2)(z-3)}=\frac{1}{z-3}-\frac{1}{z-2}=
\frac{-1}{3\left(1-\frac{z}{3}\right)}-
\frac{1}{z\left(1-\frac{2}{z}\right)}$

For $2<|z|<3$, $\ds\left|\frac{z}{3}\right|<1$ and
$\ds\left|\frac{2}{z}\right|<1$, so using the binonmial theorem
gives

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

$\ds=\cdots -\frac{2^n}{z^{n+1}}-\cdots-\frac{2}{z^2}-
\frac{1}{z}-\frac{1}{3}-\frac{z}{3^2}-\frac{z^2}{3^3}-
\cdots-\frac{z^n}{3^{n+1}}-\cdots$


\item[ii)]
$f(z)=(z-a)^kf^*(z)$ and $g(z)=(z-a)^lg^*(z)$ where $f^*$ and
$g^*$ are analytic in a neighbourhood of a and non-zero at $z=a$.

$\ds F(z)=\frac{f(z)}{g(z)}=(z-a)^{k-l}\frac{f^*(z)}{g^*(z)}$

If $k>l$, $F(z)\to0$ as $z\to a$, a removable singularity.

If $k<l$, $F(z)$ has a pole of order $l-k$ at $z=a$.

If $k=l$, $\ds
F(z)=\frac{f^*(z)}{g^*(z)}\to\frac{f^*(a)}{g^*(a)}\not=0$, so $F$
has a removable singularity.

Now if $l=1$, $g(z)=g'(a)(z-a)+\cdots$ (by Taylor's theorem)

So $g^*(a)=g'(a)$, and similarly $f^*(a)=f'(a)$.

So $\ds\lim_{z\to a}\frac{f(z)}{g(z)}=\frac{f'(a)}{g'(a)}$

\end{itemize}

\end{document}
