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I need to evaluate $\text{res}\dfrac{\sin z}{e^{z} -1}$ at all singular points. Not sure how to handle a removable singularity.

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It is possible to evaluate this in Wolfram Alpha in two steps using: Reduce[Exp[z] - 1 == 0, z] and: Table[Residue[Sin[z]/(Exp[z] - 1), {z, 2 I [Pi] n}], {n, -6, 6}] –  Mats Granvik May 11 '13 at 16:02

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up vote 4 down vote accepted

At $z=0$, as you say, the singularity is "removable," meaning that it is not really a singularity. There, the residue is zero.

For other poles at $z=i 2 \pi k$, $k \in \mathbb{Z}$, the poles are simple and the residue at these poles is simply

$$\frac{\sin{i 2 \pi k}}{e^{i 2 \pi k}} = i \sinh{2 \pi k}$$

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