TITLE: Finding $\sum_{k=0}^{n-1}\frac{\alpha_k}{2-\alpha_k}$, where $\alpha_k$ are the $n$-th roots of unity
QUESTION [10 upvotes]: The question asks to compute:
  $$\sum_{k=0}^{n-1}\dfrac{\alpha_k}{2-\alpha_k}$$
  where $\alpha_0, \alpha_1, \ldots, \alpha_{n-1}$ are the $n$-th roots of unity.

I started off by simplifiyng and got it as:
$$=-n+2\left(\sum_{k=0}^{n-1} \dfrac{1}{2-\alpha_k}\right)$$
Now I was stuck. I can rationalise the denominator, but we know $\alpha_k$ has both real and complex components, so it can't be simplified by rationalising. What else can be done?

REPLY [4 votes]: For future reference  here is a solution using residues.  We have that
with $\zeta_k = \exp(2\pi i k/n)$ so that $\zeta_k^n = 1$
$$\sum_{k=0}^{n-1} \mathrm{Res}_{z=\zeta_k}
\frac{1}{2-z} \frac{n}{z^n-1}
= \sum_{k=0}^{n-1} \left.
\frac{1}{2-z} \frac{n}{nz^{n-1}} \right|_{z=\zeta_k}
\\ = \sum_{k=0}^{n-1} \left.
\frac{1}{2-z} \frac{z}{z^{n}} \right|_{z=\zeta_k}
= \sum_{k=0}^{n-1} \frac{\zeta_k}{2-\zeta_k}$$
which is our sum $S_n.$
Now observe that
$$\mathrm{Res}_{z=2}
\frac{1}{2-z} \frac{n}{z^n-1}
= -\frac{n}{2^n-1}.$$
Furthermore the residue at infinity
$$\mathrm{Res}_{z=\infty}
\frac{1}{2-z} \frac{n}{z^n-1} = 0$$
since we have the bound $2\pi n R  / R /R^n = 2\pi n / R^n \rightarrow
0$ as $R\rightarrow\infty.$ Residues sum to zero and we get
$$S_n - \frac{n}{2^n-1} = 0
\quad\text{or}\quad
S_n = \frac{n}{2^n-1}$$
as claimed.