Series Summation: $\sum\limits_{k=1}^{N-1}\frac{1}{z-w_k}$ where $w_k=e^{\frac{2\pi i k}{N}}$ I have the series
$$\sum_{k=1}^{N-1}\frac{1}{1-w_k} $$ where $w_k=e^{\frac{2\pi i k}{N}}$, how can I find the summation of this , another question related to this sum
$$\sum_{k=1}^{N-1}\frac{1}{z-w_k}   $$
 A: Let $P(z)=z^{N}-1$. Then $$P(z)=\prod_{k=0}^{N-1}(z-w_{k})$$ where $w_{0}=1$.
From here, it is easy to check that
$$
\sum_{k=0}^{N-1}\frac{1}{z-w_{k}}=\frac{P^{\prime}(z)}{P(z)}%
$$
hence
$$
\sum_{k=1}^{N-1}\frac{1}{z-w_{k}}=\frac{P^{\prime}(z)}{P(z)}-\frac{1}{z-1}=\frac{Nz^{N-1}}{z^{N}-1}-\frac{1}{z-1}=\frac{(N-1)z^{N}-Nz^{N-1}+1}{(z^{N}-1)(z-1)}\text{.}
$$
In particular, for $z=1$:
$$
S:=\sum_{k=1}^{N-1}\frac{1}{1-w_{k}}=\lim_{z\rightarrow1}\frac{(N-1)z^{N}-Nz^{N-1}+1}{(z^{N}-1)(z-1)}.
$$
In order to compute the limit, we write is as a limit at $0$, i.e., let
$z=w+1$, $w\rightarrow0$, hence
$$
S=\lim_{w\rightarrow0}\frac{(N-1)(w+1)^{N}-N(w+1)^{N-1}+1}{w((w+1)^{N}-1)}.
$$
Next, recall that
$$
\lim_{w\rightarrow0}\frac{(w+1)^{N}-1}{w}=N
$$
hence
$$
S=\frac{1}{N}\cdot\lim_{w\rightarrow0}\frac{(N-1)(w+1)^{N}-N(w+1)^{N-1}+1}{w^{2}}.
$$
The last step is to apply l'Hopital's rule, hence
\begin{align*}
S  & =\frac{1}{N}\cdot\lim_{w\rightarrow0}\frac{\left(  (N-1)(w+1)^{N}-N(w+1)^{N-1}+1\right)  ^{\prime}}{\left(  w^{2}\right)  ^{\prime}}=\frac{1}{N}\cdot\lim
_{w\rightarrow0}\frac{N(N-1)(w+1)^{N-2}w}{2w}\\
& =\frac{1}{N}\cdot\lim_{w\rightarrow0}\frac{N(N-1)}{2}(w+1)^{N-2}=\frac{N-1}{2}.
\end{align*}
A: $$\frac{1}{1-w^{r}}+\frac{1}{1-w^{n-r}}=\frac{2-w^{r}-w^{n-r}}{2-w^{r}-w^{n-r}}=1$$ where $w=e^{i2\pi/n}$
If you continue cancelling in pairs the sum will turn out to be = $\dfrac{N-1}{2}$
In case you have a middle term ,the term = $\dfrac{1}{1-e^{i\pi/2}}=\dfrac{1}{2}$
A: The answer for the first sum is $$\sum_{k=1}^{N-1}\frac{1}{1-w_k}=\frac{N-1}{2}$$ 
And for the second:
$$\sum_{k=1}^{N-1}\frac{1}{z-w_k} = \sum_{k=0}^{N-1}\frac{1}{z-w_k}- \frac{1}{z-1}= \frac{N z^{N-1}}{z^N-1} -\frac{1}{z-1}$$
I believe series expansion in $w_k$, sum over $k$ and then summing back the series wll give you the above results.
