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

{\bf Question}

\begin{itemize}
\item[a)]
Use Cauchy's integral representation with an appropriate contour
to show that, for $|z|<1$,

$$z^n=\frac{1}{2\pi}\int_0^{2\pi}
\frac{e^{i(n+1)\theta}}{e^{i\theta}-z}d\theta$$

where $n$ is a positive integer.  What is the value of the
integral if $|z|>1$?


\item[b)]
Express the function

$$f(z)=\frac{2}{z(z-1)(z-2)}$$

in partial fractions.

Find the Laurent expansions of $f(z)$ in powers of $z$ in each of
the two regions $1<|z|<2$ and $|z|>2$.

Hence, or otherwise, evaluate $\int_C f(z)dz$ where $C$ is

\begin{itemize}
\item[i)]
the circle $\ds|z|=\frac{3}{2}$

\item[ii)]
the circle $|z|=3$.

\end{itemize}

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
For $C=$ unit circle and $|z|<1$, $\ds f(z)=\frac{1}{2\pi
i}\int_C\frac{f(w)}{w-z}dw$

so with $f(z)=z^n$ and $z=e^{i\theta}, \,\,\,
0\leq\theta\leq2\pi$,

$\ds z^n=\frac{1}{2\pi i}\int_0^{2\pi}
\frac{e^{in\theta}ie^{i\theta}d\theta}{e^{i\theta}-z}=
\frac{1}{2\pi}\int_0^{2\pi}
\frac{e^{i(n+1)\theta}d\theta}{e^{i\theta}-z}$

Now if $|z|>1$, $z$ is outside $C$ and $\ds\frac{f(w)}{w-z}$ is
analytic inside and on $C$ and so $\ds\int_C\frac{w^n}{w-z}dw=0$
by Cauchy's Theorem.


\item[b)]
$\ds f(z)=\frac{2}{z(z-1)(z-2)}=
\frac{1}{z}-\frac{2}{z-1}+\frac{1}{z-2}$

${}$

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

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

So $\ds f(z)=\cdots-\frac{z^n}{2^{n+1}}-\cdots-\frac{z^2}{8}-
\frac{z}{4}-\frac{1}{2}-\frac{1}{z}-\frac{2}{z^2}-\frac{2}{z^3}-
\cdots$

${}$

For $|z|>2$, $\ds\frac{1}{z-1}$ expands as before, but

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

So $\ds f(z)=\frac{2}{z^3}+\frac{6}{z^4}+\frac{14}{2^5}+\cdots+
\frac{2^{n-1}-2}{z^n}+\cdots$


\begin{itemize}
\item[i)]
For the circle $C_1:\, |z|=\frac{3}{2}$, this encloses $z=0$ and
$z=1$

$\ds\int_{C_1}f(z)dz=\int_{C_1}\frac{1}{z}-
2\int_{C_1}\frac{2}{z-1}+\int_{C_1}\frac{1}{z-2}$

$\hspace{0.5in}=2\pi i-2.2\pi i+0=-2\pi i$

\item[ii)]
For the circle $C_2:\, |z|=3$, this encloses $z=0$, $z=1$ and
$z=2$

$\ds\int_{C_2}f(z)dz=2\pi i-2.2\pi i+2\pi i=0$

\end{itemize}

\end{itemize}

\end{document}