PreCalc Roots of Unity Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}$$
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 $\omega, \omega^2, \omega^3, \omega^4$.
