Calculate the integral $\int_{\varGamma}\frac{3e^{z}}{1-e^{z}}dz$

Question

I am looking to solve $$\int_{\varGamma}\frac{3e^{z}}{1-e^{z}}dz,$$ where $\varGamma$ is the contour $|z|=4\pi/3$.

We have been asked first to consider $e^{z}=1$ and $e^{z}=-1$ which I get to be $z=2k\pi i$ and $(2k+1)\pi i$ so I assume this question concerns singularities but I am unsure how to start.

Answer 1

You have to compute the residue at each $ z \in \mathbb{C} : e^z=1 $. Then take the points which are in the compact component of the complement of $\Gamma$, and sum the residue with the good sign (depends of the orientation of $\Gamma$).

Answer 2

If we set $$ f(z)=\mathrm{e}^{z}-1, $$ then $$ \frac{f'(z)}{f(z)}=\frac{\mathrm{e}^{z}}{\mathrm{e}^{z}-1}, $$ and $$ \int_\Gamma\frac{3\mathrm{e}^{z}\,dz}{1-\mathrm{e}^{z}}=-3\int_{|z|=4\pi/3}\frac{f'(z)\,dz}{f(z)} $$ But $\frac{1}{2\pi i}\int_{|z|=4\pi/3}\frac{f'(z)\,dz}{f(z)}$ is the number of roots of $f(z)=\mathrm{e}^z-1=0$ in the interior of $\Gamma$. (Argument Principle.) And there is exactly one root $z=0$ is that disc.

Therefore, $$ \int_\Gamma\frac{3\mathrm{e}^{z}\,dz}{1-\mathrm{e}^{z}}=-3\int_{|z|=4\pi/3}\frac{f'(z)\,dz}{f(z)}=(-3)\cdot 2\pi i\cdot 1=-6\pi i. $$