TITLE: Sum of odd powers of even number of complex numbers
QUESTION [5 upvotes]: I’ve come across a result and I’m having a bit of trouble seeing where it comes from, and it seems like it should be obvious why it is true. Say we are given n complex numbers
$ \{z_1,...,z_n\}$ where n is even, such that the sum of their odd integer powers is zero for all powers less than $n$. Ie;
$$\sum_{i=1}^n z_i^k=0$$
for all k an odd integer less than n. Then if $m$ is an odd integer greater than n, we get that
$$\sum_{i=1}^n z_i^m=0$$
Again I have a feeling that this is a well-known result, or at least a re-wording or special example of one, but I don’t know where to start looking.

REPLY [3 votes]: This can be viewed as a consequence of the Girard-Newton identities.
If we put $p_k=\sum_{j=1}^n z_j^k$ for $k>0$, and $\prod_{j=1}^n(z-z_j)=\sum_{k=0}^{n}(-1)^k e_k z^{n-k}$ (so that $e_k$ are elementary symmetric polynomials in $z_1,\ldots,z_n$), these identities say $ke_k+\sum_{j=1}^k(-1)^j e_{k-j}p_j=0$ for $1\leqslant k\leqslant n$ (which gives $e_k=0$ for odd $k$, using induction on $k$), and $\sum_{j=k-n}^k(-1)^j e_{k-j}p_j=0$ for $k>n$ (which gives $p_k=0$ for odd $k>n$, again by induction on $k$, and we're done).
(The above basically shows that $p(z)=\prod_{j=1}^n(z-z_j)$ satisfies $p(z)=p(-z)$, so that $z\mapsto-z$ acts as a permutation on $z_1,\ldots,z_n$. This may give some intuition for other solutions.)