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Find the poles, indicate their order and compute their residues for the following functions: $$g(z)=\frac{e^z}{\sin z}$$

I have a singularity at $z=0$ where the residue would be $1$ ... however, would I also have a pole at $z=n\pi$? If yes, how would I calculate the order and residue of this?

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  • $\begingroup$ which residue theorem are you using? $\endgroup$ – JMP Apr 18 '15 at 4:06
  • $\begingroup$ for the singularity at $z=0$ i used limits $\endgroup$ – smith Apr 18 '15 at 4:07
  • $\begingroup$ i used limits for the pole at $z=n\pi$ and got the residue as $\pm exp(n\pi)$ based on $n$ being even or odd .. would this be correct? $\endgroup$ – smith Apr 18 '15 at 4:12
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There are poles at $z=n \pi$ for all integer values of $n$ since $\sin (n\pi) =0$ for all integer $n$.


To show that the poles are simple, expand $\sin z$ in the series

$$\begin{align} \sin z &=(-1)^{n} \sum_{m=0}^{\infty} (-1)^{m} \frac{(z-n\pi)^{2m+1}}{(2m+1)!}\\\\ &=(-1)^{n} (z-n\pi) \sum_{m=0}^{\infty} (-1)^{m} \frac{(z-n\pi)^{2m}}{(2m+1)!} \end{align}$$

where the last summation is not zero when $z=n\pi$.


To calculate the residues, we have

$$\lim_{z \to n \pi} (z-n\pi)\left(\frac{e^z}{\sin z}\right)=(-1)^ne^{n\pi}$$

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  • $\begingroup$ thank you, would this be considered as a simple pole? $\endgroup$ – smith Apr 18 '15 at 4:14
  • $\begingroup$ You're welcome. My pleasure. Yes, the poles are simple. $\endgroup$ – Mark Viola Apr 18 '15 at 4:15

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