$\sum z_i^k=0$ for all $k\ge 2$ implies $z_1=\dots=z_n=0$?

With a slight generalization of copper.hat's solution to

• Does $\lambda_1^n+ \lambda_2^n+ \dots +\lambda_k^n =0 $ for all $n$ imply that $\lambda_1= \lambda_2= \dots= \lambda_k = 0 $?

one can proceed as follows:

Let $p(z) = \sum_{k=0}^N a_k z^k$ be a polynomial such that $p(0) = 0$, $ p'(0) = 0$, and $$ p(z_j) = |z_j| \quad \text{for } j = 1, \ldots, n \, . $$

Then $a_0=a_1=0$ and $$ \sum_{j=1}^n |z_j| = \sum_{j=1}^n p(z_j) = \sum_{k=2}^N a_k \sum_{j=1}^n z_j^k = 0 $$ and therefore $z_1 = \ldots = z_n = 0$.

Remark 1: The existence of such a polynomial $p$ can be shown with a slightly modified Langrange interpolation: Let $w_1, \ldots, w_m$ be the non-zero distinct values in $\{ z_1, \ldots, z_n \}$, and $$ L_j(z) = \prod_{\substack{l=1 \\ l \ne j}}^m \frac{z-w_l}{w_j-w_l} \quad (1 \le j \le m) $$ the corresponding Language polynomials. Then $$ p(z) = \sum_{j=1}^m |w_j|\frac{z^2}{w_j^2} L_j(z) \, . $$ has the desired properties.

Remark 2: The degree of $p$ is at most $m + 1 \le n+1$. Therefore it suffices to require that $ z_1^k+ \ldots +z_n^k=0$ for $2 \le k \le n+1$.

Here is a proof if we assume vanishing for any consecutive sequence of $n$ exponents, $l\leq k <l+n$.

Let $w_j$, $j=1, \ldots , t$ be the distinct values amongst $z_1, \ldots , z_n$ and let $w_j$ occur $m_j>0$ many times.

Then we have the matrix equation

$$\begin{pmatrix} w_1^l&w_2^l&\cdots &w_t^l\\ w_1^{l+2}&w_2^{l+2}&\cdots &w_t^{l+2}\\ &&\vdots &\\ w_1^{l+t}&w_2^{l+t}&\cdots &w_t^{l+t}\\ \end{pmatrix}\begin{pmatrix} m_1\\ m_2\\ \vdots \\ m_t\\ \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ \vdots\\ 0\\ \end{pmatrix}$$

which implies that

$$\det \begin{pmatrix} w_1^l&w_2^l&\cdots &w_t^l\\ w_1^{l+2}&w_2^{l+2}&\cdots &w_t^{l+2}\\ &&\vdots &\\ w_1^{l+t}&w_2^{l+t}&\cdots &w_t^{l+t}\\ \end{pmatrix}=\prod\limits_j w_j^l\prod\limits_{i<j} (w_j-w_i)=0$$

Since the $w_j$ are distinct, there is at least one $j$ such that $w_j=0$. By induction the result follows.

Note that consecutivity of the exponents $k$ is essential since if $\zeta_p$ is a primitive $p$ th root of unity then

$$\sum\limits_{i=0}^{p-1} (\zeta^i)^{pn+l}=0$$ for all $l$ not a multiple of $p$.