I want to compute this integral

$$\int_{C(2i,5)} \frac{z}{e^z-1}dz$$

I was trying to trying to use residue theorem but I could not find residue of this function.

  • $\begingroup$ Why can you not find the residue? How many places in the disk does $e^z =1$??? $\endgroup$ – copper.hat Feb 19 at 21:47
  • $\begingroup$ What does $C(2i,5)$ mean? Does it denote the circle centered at $z=2i$ with radius $5$? $\endgroup$ – user170231 Feb 19 at 21:51
  • $\begingroup$ How would you proceed in general, if you needed to evaluate $\oint_{\gamma}f(z)\;dz$? $\endgroup$ – MPW Feb 19 at 21:55
  • $\begingroup$ It is simple en.wikipedia.org/wiki/Residue_(complex_analysis)#Simple_poles $\endgroup$ – copper.hat Feb 19 at 21:56
  • $\begingroup$ Yes, $C(2i,5)$ is a circle. $\endgroup$ – Grzegorz Drzyzga Feb 19 at 22:39

If $z=x+iy$ with $x,y \in \Bbb{R}$ then $e^z = e^{x+iy} = e^x e^{iy} = 1 \Leftrightarrow e^x = 1$ and $y = 2\pi k$ with $k \in \Bbb{Z}$

With this the only posible singularities in the disk are $z=0$ and $z=2\pi i$. If you evaluate $\lim_{z \rightarrow 0} \frac{z}{e^z -1} = \lim_{z \rightarrow 0} \frac{1}{e^z} = 1$, so $z=0$ is a removable singularity. Then $z= 2 \pi $ is a pole of order $1$ because

$$\lim_{z \rightarrow 2\pi i} (z- 2\pi i) \frac{z}{e^z -1} = \lim_{z \rightarrow 2\pi i} \frac{2z-2\pi i}{e^z} = 2\pi i \in \Bbb{C}^*$$

And in fact as $z= 2\pi i$ is a simple pole then the last limit it the residue in that point, so aplying the residue theorem

$$\int_{C(2i,5)} \frac{z}{e^z - 1} = 2 \pi i \; 2\pi i = -4 \pi^2$$


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