Complex numbers - roots of unity 
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
  $$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}.$$

I have tried adding the first two and the second two separately, then adding those sums but how do I get a numerical value as the answer? 
Thanks
 A: HINT : $$\frac{\omega}{1-\omega^2}+\frac{\omega^2}{1-\omega^4}+\frac{\omega^3}{1-\omega}+\frac{\omega^4}{1-\omega^3}$$
$$=\frac{\omega}{1-\omega^2}+\frac{\omega^2}{1-\omega^4}+\frac{\omega^7}{\omega^4-\omega^5}+\frac{\omega^6}{\omega^2-\omega^5}$$
A: A less efficient answer with a more general approach.

You may notice that $\{\omega,\omega^2,\omega^3,\omega^4\}$ are the roots of $\frac{x^5-1}{x-1}$. If we set $Z=\{\omega,\omega^2,\omega^3,\omega^4\}$
we have
$$ \sum_{z\in Z}\frac{z}{1-z^2}=\sum_{z\in Z}\frac{z^3}{1-z}=\sum_{z\in Z}\frac{1}{1-z}-\sum_{z\in Z}(1+z+z^2)=-2+\sum_{z\in Z}\frac{1}{1-z}.$$
If $z\in Z$, $1-z$ is a root of $\frac{1-(1-x)^5}{x}=x^4-5x^3+10x^2-10x+5$.
By Vieta's theorem it follows that
$$ \sum_{z\in Z}\frac{1}{1-z} = \frac{10}{5} = 2$$
hence:
$$ \sum_{z\in Z}\frac{z}{1-z^2} = \color{red}{0}.$$
Key steps:


*

*$z\mapsto z^3$ is a bijection on $Z$

*for any $k\in[1,4]$ we have $\sum_{z\in Z}z^k = -1$.

A: The sum of the first and fourth terms is
\begin{align*}
\frac{\omega}{1 - \omega^2} + \frac{\omega^4}{1 - \omega^3} &= \frac{\omega (1 - \omega^3) + \omega^4 (1 - \omega^2)}{(1 - \omega^2)(1 - \omega^3)} \\
&= \frac{\omega - \omega^4 + \omega^4 - \omega^6}{(1 - \omega^2)(1 - \omega^3)} \\
&= \frac{\omega - \omega^4 + \omega^4 - \omega}{(1 - \omega^2)(1 - \omega^3)} \\
&= 0,
\end{align*}and the sum of the second and third terms is
\begin{align*}
\frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} &= \frac{\omega^2 (1 - \omega) + \omega^3 (1 - \omega^4)}{(1 - \omega^4)(1 - \omega)} \\
&= \frac{\omega^2 - \omega^3 + \omega^3 - \omega^7}{(1 - \omega^4)(1 - \omega)} \\
&= \frac{\omega^2 - \omega^3 + \omega^3 - \omega^2}{(1 - \omega^4)(1 - \omega)} \\
&= 0.
\end{align*}Therefore, the sum of all four terms is $\boxed{0}$.
A: Maybe more direct
$$
\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}
=
\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^{-1}} + \frac{\omega^{3}}{1 - \omega} + \frac{\omega^{-1}}{1 - \omega^{-2}}
$$
