<s>[INST] 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!
 [/INST] 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:
$$f(\omega) = 2 \frac{(\omega^3 + \omega^2) + (\omega + \omega^4)}{(\omega^3 + \omega^2)  (\omega + \omega^4)} = 2 \frac{\omega + \omega^2 + \omega^3 + \omega^4}{\omega^3 + \omega^4 +\omega^6 + \omega^7}= 2 \frac{\omega + \omega^2 + \omega^3 + \omega^4}{\omega^3 + \omega^4 +\omega + \omega^2}=2$$
</s>