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QUESTION Find the Laurent expansions of the following functions
which converge in the regions indicated.
\begin{description}

\item[(a)]
$z^me^\frac{1}{z^2},\ 0<|z|<\infty$

\item[(b)]
$\frac{1}{(z-1)(z+2)},\ 0<|z-1|<3$

\end{description}

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ANSWER
\begin{description}

\item[(a)]
$$z^me^\frac{1}{z^2}=z^m\sum_{n=0}^\infty
\frac{z^{-2n}}{n!}=\sum_{n=0}^\infty\frac{z^{m-2n}}{n!}$$ (using
the exponential series)

\item[(b)]
\begin{eqnarray*}
\frac{1}{(z-1)(z+2)}&=&\frac{1}{z-1}\cdot\frac{1}{(z-1)+3}\\
&=&\frac{1}{z-1}\cdot\frac{1}{3}\cdot\frac{1}{1+(\frac{z-1}{3})}\\
&=&\frac{1}{3}\cdot\frac{1}{z-1}\sum_{n=0}^\infty\left(-\frac{z-1}{3}\right)^n
\textrm{ (geometric series)} \\ &=&\sum_{m=-1}^\infty
(-1)^{m+1}3^{-(m+2)}(z-1)^m\\ &&(\textrm{Taking } m=n-1)
\end{eqnarray*}

\end{description}


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