Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.