# 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

HINT : $$\frac{\omega}{1-\omega^2}+\frac{\omega^2}{1-\omega^4}+\frac{\omega^3}{1-\omega}+\frac{\omega^4}{1-\omega^3}$$ $$=\frac{\omega}{1-\omega^2}+\frac{\omega^2}{1-\omega^4}+\frac{\omega^7}{\omega^4-\omega^5}+\frac{\omega^6}{\omega^2-\omega^5}$$

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 $\boxed{0}$.

Maybe more direct $$\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3} = \frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^{-1}} + \frac{\omega^{3}}{1 - \omega} + \frac{\omega^{-1}}{1 - \omega^{-2}}$$

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:

1. $z\mapsto z^3$ is a bijection on $Z$
2. for any $k\in[1,4]$ we have $\sum_{z\in Z}z^k = -1$.