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

{\bf Question}

\begin{itemize}
\item[a)]
Find a rational function $R(z)$ having the following properties:

\begin{itemize}
\item[i)]
The only singularities are poles of order 3 at $z=+i$ and $z=-i$,
and a simple pole at $z=0$ with residue 2,

\item[ii)]
$R$ has a zero of order 2 at $z=1$,

\item[iii)]
$\ds\lim_{|z|\to\infty}z^3R(z)=1$,

\item[iv)]
$R(-1)=-1$.
\end{itemize}

What is the behaviour of $R(z)$ at infinity?


\item[b)]
Find the Laurent series for

$$f(z)=\frac{1}{(z-a)(z-b)} \hspace{0.3in} (|a|<|b|)$$

in the regions

\begin{itemize}
\item[i)]
$|a|<|z|<|b|$,

\item[ii)]
$|z|>|b|$.

\end{itemize}

\end{itemize}

\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
Let $\ds R(z)=\frac{P(z)}{Q(z)}$

Then by (i) $Q(z)=(z-i)^3(z+i)^3z=z(z^2+1)^3$

by (ii) $P(z)=(z-1)^2M(z)$

by (iii) $z^3R(z)$ has a non-zero finite limit as $|z|\to\infty$,
so $z^3P(z)$ and $Q(z)$ have the same degree, namely 7.  So $M(z)$
is quadratic.

Thus $\ds R(z)=\frac{(z-1)^2(Az^2+Bz+C)}{z(z^2+1)^3}$

$\ds\lim_{|z|\to\infty}z^3R(z)=A$ so $A=1$

res$\ds(0)=\lim_{z\to0}zR(z)=
\lim_{z\to0}\frac{(z-1)^2(z^2+Bz+C)}{(z^2+1)^3}=C=2$

Finally by (iv)

$\ds R(-1)=\frac{(-2)^2((-1)^2-B+2)}{-(2)^3}=-1$

so $3-B=2 \hspace{0.5in}$ i.e. $B=1$

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

To investigate the behaviour at infinity, consider
$R(\frac{1}{z})$

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

So $R(\frac{1}{z})$ has a zero of order 3 at $z=0$.

i.e. $R(z)$ has a zero of order 3 at $z=\infty$.


\item[b)]
$\ds{1}{(z-a)(z-b)}=
\frac{1}{a-b}\left(\frac{1}{z-a}-\frac{1}{z-b}\right)=f(z)$

Now for $|\alpha|>|z|$

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

whereas for $|\alpha|<|z|$

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

So (i) $|a|<|z|<|b|$

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

(ii) $|z|>|b|$

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

\end{itemize}

\end{document}