The sum of odd powered complex numbers equals zero implies they cancel each other in pairs Show that if a set of complex numbers $z_1,z_2,\ldots,z_n$ satisfy 
$$z_1^l+z_2^l+\cdots+z_n^l=0$$ for every odd $l$, then for any $z_i$ we can always find some $z_j$ such that $z_i+z_j=0$.
The question has been answered here for real numbers but not for complex numbers

Let $\displaystyle P(z) = \sum_{k=0}^n (-1)^ke_k z^{n-k}$ be the $n^{th}$ degree polynomial with roots $z_k \mid k=1,\cdots,n$, where by Vieta's formulas $e_k$ are the elementary symmetric polynomials.
Let $\displaystyle p_i=\sum_{k=1}^n z_k^i\,$, where it is given that $p_l=0$ for all odd $l$. 
From Newton's identities $\displaystyle k e_k = \sum_{i=1}^k (-1)^{i-1}e_{k-i}p_i$ it follows (by induction, for example) that $e_l=0$ for all odd $l$. Therefore, the polynomial $P(z)$ has every other coefficient $0$, so it contains either only even powers of $z$, or only odd powers of $z$, depending on the parity of $n$. In the first case $P(z)$ is an even function, in the second case an odd one. In both cases $P(z)=0 \iff P(-z)=0$ so the roots of $P(z)$ can be grouped in pairs of mutually opposites.
