Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find $$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}$$
So far, I have simplified that down to $\frac{-2}{(1+\omega)^2}$, but I don't know how to continue. Thank you!
Let $f(x) = x^3 / (1 + x)$. Then, you're looking for
$$ f(\omega) + f(\omega^2) + f(\omega^3) + f(\omega^4) $$
But we can simplify the formula by multiplying the top and bottom by an appropriate formula:
$$ f(x) = \frac{x^3 (1 - x + x^2 - x^3 + x^4)}{1 + x^5} $$
$$ f(x) \equiv \frac{x^3 - x^4 + 1 - x + x^2}{2} \pmod{x^5 \equiv 1} $$
Since $\omega + \omega^2 + \omega^3 + \omega^4 = -1$, we can add up the five terms of this formula separately to get
$$ f(\omega) + f(\omega^2) + f(\omega^3) + f(\omega^4) = \frac{(-1) - (-1) + 4 - (-1) + (-1)}{2} \\= 2 $$
This works because whenever $\gcd(n, 5) = 1$, the four possible values of $(\omega^i)^n$ are all four of the values $\omega, \omega^2, \omega^3, \omega^4$.
We can simplify the individual terms by reducing the numerators to $1$. For example, for the first term we have $$ \frac{\omega}{1+\omega^2} = \frac{\omega \omega^4}{(1+\omega^2)\omega^4} = \frac{1}{\omega^4 + \omega^6} = \frac{1}{\omega + \omega^4} $$ and we can do something similar with the other terms. After simplification we find $$f(\omega) = \frac{2}{\omega+\omega^4} + \frac{2}{\omega^3 + \omega^2}$$ Now it's straightforward to compute the sum of these two terms: $$f(\omega) = 2 \frac{(\omega^3 + \omega^2) + (\omega + \omega^4)}{(\omega^3 + \omega^2) (\omega + \omega^4)} = 2 \frac{\omega + \omega^2 + \omega^3 + \omega^4}{\omega^3 + \omega^4 +\omega^6 + \omega^7}= 2 \frac{\omega + \omega^2 + \omega^3 + \omega^4}{\omega^3 + \omega^4 +\omega + \omega^2}=2$$
We need $$\sum_{r=1}^4\dfrac{x^r}{1+x^{2r}}$$ where $x^5=1$
Let $y=\dfrac x{1+x^2}=\dfrac1{x+\dfrac1x}$
$$\implies\dfrac1{y^5}=\left(x+\dfrac1x\right)^5=x^5+\dfrac1{x^5}+\binom51\left(x^3+\dfrac1{x^3}\right)+\binom52\left(x+\dfrac1x\right)$$
Now as $x^3+\dfrac1{x^3}=\left(x+\dfrac1x\right)^3-3\left(x+\dfrac1x\right)=\dfrac1{y^3}-\dfrac3y$
$$\implies\dfrac1{y^5}=1+\dfrac11+5\left(\dfrac1{y^3}-\dfrac3y\right)+\dfrac{10}y$$
As $y\ne0,$ multiply both sides by $y^5$ to get $$2y^5-5y^4+\cdots=0$$
By Vieta's formula, $$\sum_{r=1}^5\dfrac{x^r}{1+x^{2r}}=\dfrac52$$
But $r=5\implies\dfrac{x^r}{1+x^{2r}}=\dfrac{x^5}{1+(x^5)^2}=\dfrac1{1+1}$ as $x^5=1$
Can you take it from here?
If $x=e^{2ir\pi/5}$ where $r\equiv0,\pm,\pm2\pmod5$
Using How to prove Euler's formula: $e^{it}=\cos t +i\sin t$?, $$x+x^{-1}=2\cos\dfrac{2r\pi}5$$ where $r\equiv0,\pm,\pm2\pmod5$
If $y=\dfrac1{2\cos\dfrac{2r\pi}5},\cos\dfrac{2r\pi}5=\dfrac1{2y}$
Using this, $\cos\dfrac{2r\pi}5,r\equiv0,\pm,\pm2\pmod5$ are the roots of $$16c^5-20c^3+5c-1=0$$
$$\implies16\left(\dfrac1{2y}\right)^5-20\left(\dfrac1{2y}\right)^3+5\left(\dfrac1{2y}\right)-1=0$$
$$\iff2y^5-5y^4+\cdots=0$$
Now follow the last part of my other answer.