How do I evaluate the Complex Integral $(z^n)/(e^z - 1)$ using residue theory?

Question

The integral I want to evaluate is the integral of $\displaystyle\frac{z^n}{e^z-1}$ over the circle $|z| = 3 \pi $. I was thinking of using Cauchy's Residue Theorem, but then I got stuck on evaluating the residues at $z=0$, because I couldn't figure out how to change the form into one where I could find a Laurent series.

Answer 1

OK, this is an attempt to rewrite my answer without mentioning L'Hôpital or Riemann, as requested.

The integrand $f(z) = \frac{z^n}{\mathrm e^z-1}$ is the quotient of two holomorphic functions, each of which we can write as a power series at any point $z_0$. For $z_0=0$, this yields

$$f(z)=\frac{z^n}{\left(\sum_{k=0}^\infty z^k/k!\right)-1}=\frac{z^n}{\sum_{k=1}^\infty z^k/k!}=\frac{z^n}{z\sum_{k=1}^\infty z^{k-1}/k!}=\frac{z^n}{z\left(1+\sum_{k=2}^\infty z^{k-1}/k!\right)}\;.$$

If $n$ is a positive integer, we can cancel one factor $z$ to obtain

$$f(z)=\frac{z^{n-1}}{1+\sum_{k=2}^\infty z^{k-1}/k!}\;.$$

The denominator doesn't go to zero as $z\to0$, and thus there is no pole at $z=0$.

On the other hand, if $n=0$, we have

$$f(z)=\frac{1}{z\left(1+\sum_{k=2}^\infty z^{k-1}/k!\right)}\;.$$

Thus we can determine

$$\lim_{z\to0}zf(z) = \lim_{z\to0}\frac{1}{\left(1+\sum_{k=2}^\infty z^{k-1}/k!\right)}=1\;,$$

and this is the residue of $f$ at $z=0$.

The residues at $z=\pm2\pi\mathrm i$ can be determined similarly using $\mathrm e^{z}=\mathrm e^{z\mp2\pi\mathrm i}$.