sum of the residues of all the isolated simgualrities Prove that, for $n \geq 3$, the sum of the residues of all the isolated singularities of 
$$\frac{z^n}{1+z+z^2+\cdots+z^{n-1}}$$
is 0
Can someone show me how to do this problem. Thank you. 

Let $\omega=e^{2\pi i/n}$.  For $k=1,2,\ldots,n-1$, the function has a simple pole at $\omega^k$ and the residue is
$$\eqalign{r_k
  &=\lim_{z\to\omega^k}\frac{(z-\omega^k)z^n}{1+z+\cdots+z^{n-1}}\cr
  &=\lim_{z\to\omega^k}\frac{(z-1)(z-\omega^k)z^n}{z^n-1}\cr
  &=\lim_{z\to\omega^k}\frac{(2z-1-\omega^k)z^n+(z-1)(z-\omega^k)z^{n-1}}{nz^{n-1}}\qquad\hbox{(L'Hopital)}\cr
  &=\frac1n\lim_{z\to\omega^k}\bigl((2z-1-\omega^k)z+(z-1)(z-\omega^k)\bigr)\cr
  &=\frac{(\omega^k-1)\omega^k}{n}\ .\cr}$$
The sum of the residues can be obtained by adding up two geometric series:
$$\sum_{k=1}^{n-1}r_k=\sum_{k=0}^{n-1}\frac{\omega^{2k}-\omega^k}{n}
  =\frac1n\Bigl(\frac{\omega^{2n}-1}{\omega^2-1}-\frac{\omega^n-1}{\omega-1}\Bigr)=0\ .$$
Notes.


*

*I have started the sum at $k=0$ instead of $k=1$: this makes the summation easier, and doesn't change the result because the extra term I've added is zero.

*The denominator $\omega^2-1$ is not zero because $n\ge3$.

