let $x_1, x_2, ... , x_{2014} $ be the roots for $x^{2014} + x^{2013} + ... +x +1 = 0 $ find $\sum_{k=1}^{2014} 1/(1-x_k) $ let $x_1, x_2, ... , x_{2014} $ be the roots for $x^{2014} + x^{2013} + ... +x +1 = 0 $ find $\sum_{k=1}^{2014} 1/(1-x_k) $.
So What I tried to do is to consider x^3 function and get the elementary symmetric functions. the find the sum of  $\sum_{k=1}^{3} 1/(1-x_k) $. I lost and did not work.
How I can approach this problem?
 A: Let $y_k=\dfrac1{1-x_k}\implies x_k=\dfrac{y_k-1}{y_k}$ and we need
$$\sum_{k=1}^{2014}y_k$$
Replacing $x$ with $\dfrac{y-1}y$ in the given equation to get
$$0=\sum_{r=0}^{2014}y^r(y-1)^{2014-r}$$
$$2015y^{2014}-y^{2013}\sum_{k=0}^{2014}\binom k1+\cdots+1=0$$
Apply Vieta's formula
A: We notice that the equation:
$$x^{2014} + x^{2013} + ... +x +1 = 0 $$
is a geometric sequence and can be rewritten as:
$$x^{2014} + x^{2013} + ... +x +1 = \sum_{n=0}^{2014}x^n = \frac{1-x^{2015}}{1-x} = 0 $$
From the denominator $x-1 $ we conclude that $x\ne1$, but we can rewrite $ 1 = e^{2\pi k} $. So from:
$$ 1-x^{2015} = 0 \rightarrow x^{2015} = e^{2\pi k} $$
$$ x_k = e^{2\pi k/2015} $$
Where $k \ne 0$ and goes from $1$ to $2014$. Because $k\ne 0$ then $x\ne1$ and there is no problem with the denominator anymore. The sum becomes:
$$ \sum_{k=1}^{2014} \frac{1}{1-x_k} = \sum_{k=1}^{2014} \frac{1}{1-e^{2\pi k/2015}} = \sum_{k=1}^{2014} \frac{e^{-\pi k/2015}}{e^{-\pi k/2015}-e^{\pi k/2015}} = \sum_{k=1}^{2014} \frac{\cos(\pi k/2015)-i\sin(\pi k/2015)}{-2i\sin(\pi k/2015)} = $$
$$ \sum_{k=1}^{2014} \frac{1}{2} + i\frac{\cot(\pi k/2015)}{2} = 1007 + \frac{i}{2}\sum_{k=1}^{2014}\cot(\pi k/2015) = 1007 $$
From Wolfram alpha $ \sum_{k=1}^{2014}\cot(\pi k/2015) = 0 $ but you can prove it easily. An example:
$$ \cot(\frac{2014\pi}{2015}) = \cot(\frac{2015\pi-\pi}{2015}) = \cot(\pi- \frac{\pi}{2015}) = -\cot(\frac{\pi}{2015})  $$
which will cancel out with the first term and so on with other terms.
