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QUESTION

Find the singularities of the following functions and work out the
residues at these points.
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\item[(a)]
$\frac{1}{z^4+4z^2}$

\item[(b)]
$\frac{1}{z^2\sin z}$

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ANSWER
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\item[(a)]
$\displaystyle\frac{1}{z^4+4x^2}=\frac{1}{z^2(z^2+4)}$.

This has a double pole at $z=0$ and simple poles at $z=\pm 2i$\\
Res(0)=$\lim_{z\to 0}\frac{d}{dz}\frac{1}{z^2+4}=\lim_{z\to
0}-\frac{2z}{(z^2+4)^2}=0$\\
Res$(2i)=\lim_{z\to2i}\frac{1}{z^2(z+2i)}=\frac{1}{-4\cdot4i}=\frac{i}{16}$\\
Res$(-2i)=-\frac{i}{16}$

\item[(b)]
$\frac{1}{z^2 \sin
z}=\frac{1}{z^2(z-\frac{z^3}{3!}+\ldots)}=\frac{1}{z^3(1-\frac{z^2}{3!}+\ldots)}
=z^{-3}(1+\frac{z^2}{3!}+\ldots)=z^{-3}+\frac{1}{3!}z^{-1}$\\
$z=0$ is a triple pole with residue $\frac{1}{3!}.$

$ z=n\pi, n\in \textbf{Z}$ are simple poles (except for $z=0$)\\
$\frac{1}{z^2 \sin z}\approx \frac{1}{(n \pi)^2(\cos
n\pi)(z-n\pi)}\Rightarrow \textrm{ Res }=\frac{(-1)^n}{(n\pi)^2}$


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