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. 
 A: It's not clear whether "all the isolated singularities" is meant to include $\infty$ (i.e. in the Riemann sphere), or just those in $\mathbb C$.
However, it turns out not to matter.
For any rational function, the sum of the residues at all poles (including $\infty$) is $0$.  In this case, the residue at $\infty$ is $0$, since
$$\dfrac{z^n}{1+z+\ldots + z^{n-1}} = \dfrac{z^n (z-1)}{z^n-1} = \dfrac{z-1}{1-1/z^n} = z - 1 + O(z^{1-n}) \ \text{as}\ |z| \to \infty$$
Thus the sum of the residues, whether you include $\infty$ or not, is $0$.
A: 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
$$
S_n=\frac1n\left[\frac{1-(z_n^n)^2}{1-z_n^2}-\frac{1-z_n^n}{1-z_n}\right]=0.
$$
A: 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
$$\sum_{j=1}^{n-1}\alpha_j = \lim_{|z|\to\infty}z \left(\sum_{j=1}^{n-1}\frac{\alpha_j}{z - \lambda_j}\right)
= \lim_{|z|\to\infty} \frac{z(z-1)}{z^n-1} = 0 \quad\text{ when }\quad n > 3.
$$
A: 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 $n\ge3$.

