$\sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}$ How can I go about computing
$$
\sum_{k\ =\ 1}^{n - 1}
\left(1 - \mathrm{e}^{\large 2\pi k\mathrm{i}/n}\right)^{-1}\
{\Large ?}
$$
I originally thought that it was supposed to be the reciprocal of the sum, and I ended up with $1/n$, but now I realized that it is the sum of the reciprocals. I've tried using
$$e^{ix}=\cos x+i\sin x,$$
but I didn't get anywhere with that.
 A: The answer is $\frac{n-1}{2}$. There are multiple ways to approach it, here is one way;
Let $\displaystyle \alpha_k = e^{i \frac{2\pi k}{n}}$ for $k = 1, \dots,  n-1$. We seek to evaluate,
$$\sum_{k=1}^{n-1} \frac{1}{1-\alpha_k}, \ \ (*) $$
We know that $\alpha_1, \dots , \alpha_{n-1}$ are roots of the polynomial,
$$P(z) = 1 + z+ \dots + z^{n-1}$$
This is because $1, \alpha, \dots, \alpha_{n-1}$ are the roots of the polynomial $1 - z^n$ (roots of unity), so when we take away $1$ as a root we obtain $P$. But $\alpha_k$ is a root of $P$ simply means that $P(\alpha_k) = 0$ for each $k$. Our goal now is to find a polynomial that has roots, 
$$\frac{1}{1-\alpha_k} = f(\alpha_k) $$
as roots instead of $\alpha_k$, with this polynomial, we find the sum of the roots of that polynomial to obtain $(*)$. Since $f$ is one-to-one we can find the inverse function,
$$f^{-1}(x) = 1 - \frac{1}{x} \Rightarrow f^{-1} \left(\frac{1}{1-\alpha_k} \right) = \alpha_k$$
Therefore the function,
$$Q(x) = P(f^{-1}(x)) = P \left(1 - \frac{1}{x}\right) $$
has roots $\displaystyle \frac{1}{1-\alpha_k}$ for each $k$ (this is the crux of the argument, please convince yourself of this!). Note that $Q$ as defined is not a polynomial yet, but we can use $Q$ as a function to find a polynomial that has the same roots. 
Now, we have that,
$$Q(x) = P \left(1 - \frac{1}{x} \right) = 1 + \left(1 - \frac{1}{x} \right) + \dots + \left( 1- \frac{1}{x} \right)^{n-1} $$
Summing this geometric series we obtain,
$$Q(x) = x\left(1 -  \left(1-\frac{1}{x} \right)^n\right) $$
With some manipulation we obtain that,
$$R(x) = x^{n-1}Q(x) = x^n - (x-1)^n = nx^{n-1} - \frac{n(n-1)}{2} x^{n-2} + \dots $$
Now we have a polynomial $R$ and I claim that $R$ has roots $\displaystyle \frac{1}{1-\alpha_k}$ for each $k$. This is because,
$$R \left(\frac{1}{1-\alpha_k} \right) = \frac{1}{(1-\alpha_k)^{n-1}} Q \left(\frac{1}{1-\alpha_k} \right) =\frac{1}{(1-\alpha_k)^{n-1}} P\left(f^{-1}\left(\frac{1}{1-\alpha_k} \right) \right) = \frac{1}{(1-\alpha_k)^{n-1}} P(\alpha_k) = 0 $$
Therefore to find $(*)$ we note that it is simply the sum of the roots of $R$, therefore,
$$ \sum_{k=1}^{n-1} \frac{1}{1-\alpha_k} = \frac{n(n-1)/2}{n} = \frac{n-1}{2}$$
This completes the proof.
A: $\sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}$
I will show that
the sum is
$\dfrac{n-1}{2}
$.
This is undoubtedly well-known,
but it wasn't to me until
I did this.
Using your suggestion,
since
$\frac1{a+bi}
=\frac{a-bi}{a^2+b^2}
$,
$\begin{array}\\
\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}
&=\left(1-\cos(2\pi k /n)-i\sin(2\pi k/n)\right)^{-1}\\
&=\dfrac{1-\cos(2\pi k /n)+i\sin(2\pi k/n)}{(1-\cos(2\pi k /n))^2+\sin^2(2\pi k/n)}\\
&=\dfrac{1-\cos(2\pi k /n)+i\sin(2\pi k/n)}{1-2\cos(2\pi k /n)+\cos(2\pi k /n)^2+\sin^2(2\pi k/n)}\\
&=\dfrac{1-\cos(2\pi k /n)+i\sin(2\pi k/n)}{2-2\cos(2\pi k /n)}\\
&=\dfrac{1-\cos(2\pi k /n)+i\sin(2\pi k/n)}{2(1-\cos(2\pi k /n))}\\
&=\dfrac12+i\dfrac{\sin(2\pi k/n)}{2(1-\cos(2\pi k /n))}\\
&=\dfrac12+i\dfrac{2\sin(\pi k/n)\cos(\pi k/n)}{2(2\sin^2(\pi k /n)))}\\
&=\dfrac12+\dfrac{i}{2}\dfrac{\cos(\pi k/n)}{\sin(\pi k /n)}\\
&=\dfrac12+\dfrac{i}{2}\cot(\pi k/n)\\
\end{array}
$
so that
$\begin{array}\\
\sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}
&=\sum_{k=1}^{n-1}\left(\dfrac12+\dfrac{i}{2}\cot(\pi k/n) \right)\\
&=\dfrac{n-1}{2}+\dfrac{i}{2}\sum_{k=1}^{n-1}\cot(\pi k/n) \\
\end{array}
$
However,
if
$S
=\sum_{k=1}^{n-1}\cot(\pi k/n)
$,
then
$S
=\sum_{k=1}^{n-1}\cot(\pi (n-k)/n)
=\sum_{k=1}^{n-1}\cot(\pi-\pi k/n)
$,
so that
$2S
=\sum_{k=1}^{n-1}(\cot(\pi k/n)+\cot(\pi-\pi k/n))
=0
$
since
$\begin{array}\\
\cot(x)+\cot(\pi-x)
&=\dfrac{\cos(x)}{\sin(x)}+\dfrac{\cos(\pi-x)}{\sin(\pi-x)}\\
&=\dfrac{\cos(x)}{\sin(x)}+\dfrac{-\cos(x)}{\sin(x)}\\
&=0\\
\end{array}
$
A: With $\zeta = \exp(2\pi i/n)$ and
$$f(z) = \frac{1}{1-z} \frac{n/z}{z^n-1}$$
we have for $1\le k\le n-1$
$$\mathrm{Res}_{z=\zeta^k} f(z)
= \frac{1}{1-\zeta^k}$$
so that
$$S = \sum_{k=1}^{n-1} \frac{1}{1-\zeta^k}
= \sum_{k=1}^{n-1} \mathrm{Res}_{z=\zeta^k} f(z).$$
Residues sum to zero and the residue at infinity is zero, so we find
$$S = - \mathrm{Res}_{z=1} f(z) - \mathrm{Res}_{z=0} f(z).$$
For the first one we have
$$- \mathrm{Res}_{z=1} f(z)
= \mathrm{Res}_{z=1} \frac{1}{z-1} \frac{n/z}{z^n-1}
\\ = \mathrm{Res}_{z=1} \frac{1}{(z-1)^2}
\frac{n/z}{1+z+\cdots+z^{n-1}}
\\ = n
\left.\left(\frac{1}{z} \frac{1}{1+z+\cdots+z^{n-1}}
\right)'\right|_{z=1}
\\ = n
\left.\left(- \frac{1}{z^2} \frac{1}{1+z+\cdots+z^{n-1}}
- \frac{1}{z} \frac{(1+\cdots+(n-1) z^{n-2})}
{(1+z+\cdots+z^{n-1})^2}
\right)\right|_{z=1}
\\ = n
\left( - \frac{1}{n} - \frac{1}{n^2} \frac{1}{2} (n-1) n \right)
= -1 - \frac{1}{2} (n-1).$$
The second one is
$$- \mathrm{Res}_{z=0} f(z) = - (1 \times n \times -1) = n.$$
Collecting everything we get
$$\bbox[5px,border:2px solid #00A000]{
S = \frac{1}{2} (n-1).}$$
A: Call $\omega = \exp(2\pi\text i/n)$. You have
$$
\sum_{k=1}^{n-1} \frac  1{1-\omega^k} = 
\sum_{k=1}^{n-1} \frac  1{1-\omega^{n-k}}\\
\implies 
\sum_{k=1}^{n-1} \frac  1{1-\omega^k}
=
\frac 12 \sum_{k=1}^{n-1} \frac  1{1-\omega^k} + \frac 1{1-\omega^{n-k}}\\
= \frac 12 \sum_{k=1}^{n-1} \frac  {2-\omega^k-\omega^{n-k}}{2-\omega^k-\omega^{n-k}} =\frac{n-1}2
$$
