Finding $\sum_{k=0}^{n-1}\frac{\alpha_k}{2-\alpha_k}$, where $\alpha_k$ are the $n$-th roots of unity 
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?
 A: I think the following answers the question using the method that Jyrki posted here.
Since $\alpha_k$ are nth roots, so they satisfy $$f(x)=x^n-1=\prod_{k=0}^{n-1}(x-\alpha_k)$$
Putting in logarithm and derivating,
$$f'(x)/f(x)=\dfrac{nx^{n-1}}{x^n-1}=\sum_{k=0}^{n-1}\dfrac{1}{x-\alpha_k}$$
Thus $$f'(2)/f(2) = \sum_{k=0}^{n-1}\dfrac{1}{2-\alpha_k} = \dfrac{n\cdot 2^{n-1}}{2^n-1}$$
Thus the required answer is given as:
$$-n+ 2\left(\dfrac{n\cdot 2^{n-1}}{2^n-1}\right)$$
$$=\dfrac{n}{2^n-1}$$
A: 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.
A: Since $\alpha_0,\alpha_1,\alpha_2, \dots , \alpha_{n-1}$ are roots of the equation
$$x^n-1=0 ~~~~~~~~~~~~~ \cdots ~(1)$$ 
You can apply Transformation of Roots to find a equation whose roots are$$\frac{1}{2-\alpha_0} , \frac{1}{2-\alpha_1},\frac{1}{2-\alpha_2}, \dots , \frac{1}{2-\alpha_{n-1}}$$
Let $P(y)$ represent the polynomial whose roots are $\frac{1}{2-\alpha_k}$
$$y=\frac{1}{2-\alpha_k}=\frac{1}{2-x} \implies x=\frac{2y-1}{y}$$
Put in $(1)$
$$\Bigg(\frac{2y-1}{y}\Bigg)^n-1=0 \implies (2y-1)^{n}-y^{n}=0$$
Use Binomial Theorem to find coefficient of $y^n$ and $y^{n-1}$.You will get sum of the roots using Vieta's Formulas.
Hope it helps!
