Given unit roots $ω^5=1$, $ω≠1$, find $\frac ω{1+ω^2} + \frac{ω^2}{1+ω^4} + \frac{ω^3}{1+ω} + \frac{ω^4}{1+ω^3}$

Question

Let ω be a complex number such that $ω^5=1$ and $ω≠1$. Find $$\frac ω{1+ω^2} + \frac{ω^2}{1+ω^4} + \frac{ω^3}{1+ω} + \frac{ω^4}{1+ω^3}$$

I've been having trouble with this unit, need help with solving this problem. (Sorry, I don't know how to format it correctly.)

Answer 1

Note that $$\frac ω{1+ω^2}\cdot\frac{ω^3}{ω^3} = \frac{ω^4}{1+ω^3}, \>\>\>\>\>\>\> \frac{ω^3}{1+ω}\cdot\frac{ω^4}{ω^4} = \frac{ω^2}{1+ω^4}$$

Thus

$$\begin{aligned} & \frac ω{1+ω^2} + \frac{ω^2}{1+ω^4} + \frac{ω^3}{1+ω} + \frac{ω^4}{1+ω^3} \\ = &\ 2\left(\frac ω{1+ω^2} + \frac{ω^3}{1+ω} \right) =2\cdot \frac{ω(1+ω)+ω^3(1+ω^2)}{(1+ω^2)(1+ω)}\\ =&\ 2\cdot \frac{ω+ω^2+ω^3+1}{1+ω^2+ω+ω^3}=2 \end{aligned}$$

Answer 2

I thought of a really symmetric way to attack this sum: Think in terms of sum and difference of fifth powers: $$1\pm\omega^{5j}=(1\pm\omega^j)\sum_{k=0}^4(\mp\omega^j)^{k}$$ From the lower signs we conclude that, for $1\le j\le9$ $$\sum_{k=1}^4\omega^{jk}=-1+5\delta_{j5}$$ While from the upper signs we get $$\begin{align}\sum_{k=1}^4\frac{\omega^k}{1+\omega^{2k}}&=\sum_{k=1}^4\omega^k\cdot\frac12\sum_{j=0}^4(-1)^j\omega^{2jk}=\frac12\sum_{j=0}^4(-1)^j\sum_{k=1}^4\omega^{(2j+1)k}\\ &=\frac12\sum_{j=0}^4(-1)^j(-1+5\delta_{j2})=\frac12(-1+5)=2\end{align}$$