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 $\lambda_j, j = 1,\ldots,n-1$ be $j^{th}$ root of $1 + z+ \cdots + z^{n-1}$ and $\alpha_j$ be the corresponding residue. The key is $\lambda_j$ are all distinct. We have following partial fraction decomposition:
$$\frac{z^n}{1+z+\cdots+z^{n-1}} = \frac{z^n}{z^n - 1}(z-1)
= z - 1 + \frac{z-1}{z^n-1} 
= z - 1 + \sum_{j=1}^{n-1}\frac{\alpha_j}{z - \lambda_j}$$
This implies
$$\sum_{j=1}^{n-1}\frac{\alpha_j}{z - \lambda_j} = \frac{z-1}{z^n-1}$$
As a consequence, we can evaluate the sum of residues as
$$\sum_{j=1}^{n-1}\alpha_j = \lim_{|z|\to\infty}z \left(\sum_{j=1}^{n-1}\frac{\alpha_j}{z - \lambda_j}\right)
= \lim_{|z|\to\infty} \frac{z(z-1)}{z^n-1} = 0 \quad\text{ when }\quad n > 3.
$$
