# Compute the integral using complex analysis method

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.

• Why can you not find the residue? How many places in the disk does $e^z =1$??? – copper.hat Feb 19 at 21:47
• What does $C(2i,5)$ mean? Does it denote the circle centered at $z=2i$ with radius $5$? – user170231 Feb 19 at 21:51
• How would you proceed in general, if you needed to evaluate $\oint_{\gamma}f(z)\;dz$? – MPW Feb 19 at 21:55
• – copper.hat Feb 19 at 21:56
• Yes, $C(2i,5)$ is a circle. – 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
$$\int_{C(2i,5)} \frac{z}{e^z - 1} = 2 \pi i \; 2\pi i = -4 \pi^2$$