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\begin{document}
\newcommand\ds{\displaystyle}
\parindent=0pt

QUESTION

Find the residues at each of the singularities of each of the
following functions, stating the type of singularity in each case
and, if a pole, its order:

$$\textrm{(i)}\ \frac{1}{z-6i}\ \ \textrm{(ii)}\ \
\frac{z^2}{z^3-1}\ \ \textrm{(iii)}\ \ \frac{1}{(z^2+4)^2}\ \
\textrm{(iv)}\ \ \frac{1}{z}e^\frac{1}{z^2}$$



\bigskip

ANSWER
\begin{description}

\item[(i)]
 $\ds\frac{1}{z-6i}:z=6i$ is a simple pole, Res = 1

\item[(ii)]
 $\ds\frac{z^2}{z^3-1}:z=1,\ e^{\frac{2 \pi i}{3}},\ e^{\frac{4 \pi
 i}{3}}$ are simple poles.
 Res $\ds=\lim_{z\to
 a_i}\frac{z^2}{3z^2}=\frac{1}{3}$ for each pole by l'H\^{o}pital.

\item[(iii)]
 $\ds\frac{1}{(z^2+4)^2}$, $z=\pm 2i$ are double poles.
 \begin{eqnarray*}
 &amp; &amp;\textrm{Res} \left( \frac{1}{(z+2i)^2(z-2i)^2} ,2i \right)\\
 &amp; = &amp; \lim_{z \to 2i}\frac{d}{dz}\frac{1}{(z+2i)^2}\\
 &amp; = &amp; \lim_{z \to 2i}-\frac{2}{(z+2i)^3}\\
 &amp; = &amp; -\frac{i}{32}
 \end{eqnarray*}
 $$\textrm{Res}\left(\frac{1}{(z+2i)^2(z-2i)^2},-2i\right)=\frac{i}{32}$$

\item[(iv)]
 $\ds\frac{1}{z}e^\frac{1}{z^2}:z=0$ is an essential singularity.\\
 $\ds\frac{1}{z}e^\frac{1}{z^2}=\frac{1}{z}
 \left(1+\frac{1}{z^2}+\frac{1}{2!}\frac{1}{z^4}+\ldots\right)\Rightarrow
 $Res = 1

\end{description}

\end{document}
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