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Use Residue theorem to compute contour integral $$\int_C \frac{4e^z}{\sin z} dz$$

I need help figuring out singularities that are within the circle $|z|= 4$. I am stuck at that part. Thanks in advance.

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  • $\begingroup$ Do you know how to look for singularities of a function of the form $h(x)=f(x)/g(x)$? $\endgroup$ – Harto Saarinen Nov 27 '14 at 21:50
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Since $\sin(n\pi)=0$ and $e^{n\pi}\neq0$, $n=0,\pm1,\pm2,...$ $z=n\pi$ and $\lim_{z->n\pi}\frac{e^z}{\sin(z)}=\infty$, $z=n\pi$ are simple poles of $\frac{e^z}{\sin(z)}$. $z=-\pi,0,\pi$ are inside the circle $|z|=4$. Then by residue theorem

$\int_{|z|=4}\frac{e^z}{\sin(z)}dz=2\pi i(Res_{z=-\pi}f(z))+Res_{z=0}f(z))+Res_{z=\pi}f(z))=2\pi i(1-e^{\pi}-e^{-\pi})=2\pi i(1-2\cos(\pi))=6\pi i$.

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