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
$$
F(z)=\frac{z^n}{1+z+z^2+\ldots+z^{n-1}}=\frac{P(z)}{Q(z)}.
$$
Since $Q(1)=n\ne 0$, then, for every $z\ne 1$ we have
$$
Q(z)=\frac{1-z^n}{1-z},
$$
and $F$ can be redefined as
$$
F(z)=\begin{cases}
\frac{(z-1)z^n}{z^n-1} &\mbox{ for } z\ne 1\\
\frac1n &\mbox{ for } z=1
\end{cases}
$$
Therefore, the set of isolated singularities of $F_n$ is given by:
$$
Q^{-1}(0)=\{z_{k,n}=z_n^k:\, 1\le k\le n-1\},\quad z_n=e^{i\frac{2\pi}{n}}
$$
We want to calculate the sum
$$
S_n=\sum_{k=1}^{n-1}\mathrm{Res}(F_n,z_n^k),
$$
where
$$
\mathrm{Res}(F,z_n^k)=\frac{(z_n^k-1)(z_n^k)^n}{n(z_n^k)^{n-1}}=\frac{(z_n^k)^2-z_n^k}{n},
$$
and we should assume that $n\ge 3$, because for $n=2$ the set $Q^{-1}(0)$ contains one element.
We get:
\begin{eqnarray}
nS_n&=&\sum_{k=1}^{n-1}\left[(z_n^k)^2-z_n^k\right]=\sum_{k=1}^{n-1}(z_n^2)^k-\sum_{k=1}^{n-1}z_n^k=\sum_{k=0}^{n-1}(z_n^2)^k-\sum_{k=0}^{n-1}z_n^k\\
&=&\frac{1-(z_n^2)^n}{1-z_n^2}-\frac{1-z_n^n}{1-z_n}=\frac{1-(z_n^n)^2}{1-z_n^2}-\frac{1-z_n^n}{1-z_n}.
\end{eqnarray}
Using the fact that $z_n^n=1$, we conclude that
$$
S_n=\frac1n\left[\frac{1-(z_n^n)^2}{1-z_n^2}-\frac{1-z_n^n}{1-z_n}\right]=0.
$$
