Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega\neq 1$, find ... 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've been having trouble with this unit, need help  on solving this problem.
 A: Hint: using $\omega^5=1$ and bringing all fractions to numerator $1$ gives:
$$
\require{cancel}
\begin{align}
\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3} & = \frac{\omega^4}{\omega^4}\frac{\omega}{1 - \omega^2} + \frac{\omega^3}{\omega^3}\frac{\omega^2}{1 - \omega^4} + \frac{\omega^2}{\omega^2}\frac{\omega^3}{1 - \omega} + \frac{\omega}{\omega}\frac{\omega^4}{1 - \omega^3} = \\
 & = \cancel{\frac{1}{\omega^4-\omega}} + \bcancel{\frac{1}{\omega^3-\omega^2}}+\bcancel{\frac{1}{\omega^2-\omega^3}} + \cancel{\frac{1}{\omega-\omega^4}}
\end{align}
$$
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: $\omega$ is what is called a root of unity. By looking at $e^{i\theta}$ we see that it must be of the form $e^{\frac{2\pi i}{5}k}$ for some $k\in\{1,2,3,4\}$. Playing around with this form reveals a bunch of interesting properties. Notice that the successive powers of $\omega$ give all four numbers of this form, and then give $1$ as the fifth power. This means that the powers of $\omega$ act like the integers modulo $5$.
You can also quickly observe that $1+\omega+\omega^2+\omega^3+\omega^4=0$
See if you can use these facts to simplify the expression some!
A: 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 $\sum_{z\in Z}z^k = -1$.

