Construct a sequence of complex numbers $z_n$ such that $z_1=1$ and $S:=\displaystyle\sum_{n=1}^\infty z_n^k = 0$ for every $k \in \mathbb{N}$.
I have some preliminary ideas. For now, fix $k=1$. Let $S_{m-1} = \displaystyle\sum_{n=1}^{m-1} z_n$ be the $m-1^{st}$ partial sum of $S$. Then $z_1 = 1$ then we want $1+ z_2 + \dots +z_{m-1} =0$ to be true. Then in this case, choosing the $m^{th}$ roots of unity $\{1,\zeta, \zeta^2, \dots, \zeta^{m-1}\}$ $($where $\zeta$ is a primitive $m^{th}$ root of unity$)$ will work here since
$$1 + \zeta + \zeta^2 + \dots +\zeta^{m-1} = 0$$
So this choice of $z_n$ works for finite sums and when $k=1$. But exploring beyond $k=1$ becomes treacherous because for example if we consider the third roots of unity $\{1,e^{i 2\pi/3}, e^{i 4\pi/3} \}$, then the choice of $k=3$, i.e. the map $z \mapsto z^3$ smashes all of them into $1=e^{2\pi i} = e^{4\pi i}$. The problem here is that the denominator of the arguments remain static while the numerators grow through the choice of $k$. So we need to modify the roots of unity so that the denominator grows rapidly enough that the map $z \mapsto z^k$ doesn't ruin the integrity of the uniform distribution that roots of unity enjoy.
One idea I had involves instead working with something that looks something like this: $z_n = e^{2\pi i n /n!}$. Then as the summation runs over $n=1, \dots, m-1$, the numerator still increases but the denominator grows much quicker than the numerator. And for $k \geq 2$, $z_n^k = e^{2\pi i k n/n! }$ still has denominator which continues to grow much quicker than the numerator. The question is $-$ does this choice of the denominator grow too quickly?
What advice may be offered regarding how to now tackle this for any value $k$ and for infinite sum $S = \displaystyle\lim_{m \to \infty} \sum_{n=1}^m z_n$ ?
