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

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 $\boxed{0}$.
