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QUESTION

Find the principal part of the Laurent expansions of the following
functions at the point $a.$
\begin{description}

\item[(a)]
$\frac{e^z}{\left(z-2\right)^2},\ a=2$

\item[(b)]
$\frac{\sin z}{z^3\left(z+1\right)^2},\ a=0$

\end{description}

\bigskip

ANSWER

\begin{description}

\item[(a)]
\begin{eqnarray*}
\frac{e^z}{\left(z-2\right)^2}&amp;=&amp;\left(z-2\right)^{-2}e^2e^{z-2}\\
&amp;=&amp;e^2\sum_{n=0}^\infty \frac{\left(z-2\right)^{n-2}}{n!} \\
&amp;=&amp;\sum_{m=-2}^\infty
\frac{e^2}{\left(m+2\right)!}\left(z-2\right)^m
\left(\textrm{Taking } m=n-2\right)
\end{eqnarray*}

 The principal part is

$e^2\left(z-2\right)^{-2}+e^2\left(z-2\right)^{-1}$

\item[(b)]
\begin{eqnarray*}
\frac{\sin
z}{z^3\left(z+1\right)^2}&amp;=&amp;z^{-3}\left(z-\frac{z^3}{3!}+\ldots\right)\left(1-z+z^2-\ldots\right)^2\\
&amp;=&amp;z^{-2}\left(1-\frac{z^2}{6}+\ldots\right)\left(1-2z+3z^2+\ldots\right)\\
&amp;=&amp;z^{-2}\left(1-2z+\left(3-\frac{1}{6}\right)z^2+\ldots\right)\\
&amp;=&amp;z^{-2}-2z^{-1}+\left(3-\frac{1}{6}\right)+\ldots
\end{eqnarray*}

The principal part is $z^{-2}-2z^{-1}$

\end{description}


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