Find $\sum_{k=1}^n P(z_k)$ where P is a polynomial and $z_k$ the $k$th root of $z^{13}=1$ Question: Given a complex number z such that $z^{13}=1$, find the sum of all possible values of $z+z^3+z^4+z^9+z^{10}+z^{12}$.
I know we have to use roots of unity and try to manipulate the polynomial we want to evaluate, but can't find a pattern.

Since $13$ is prime, the map $z\mapsto z^k$ is a permutation of the roots of $z^{13}-1$, for $k=1,2,3,\dots,12$.  Thus, the answer is $6$ times the sum of the roots , and this is $0$ by Vieta.
