It's known that sum of all $n$'th roots of some $z \in \mathbb C$ with $|z| = 1$ is zero (if $n \geqslant 2$).
Is it true that any zero sum of roots of unity can be decomposed in this way? That is if we have $\zeta_1 + \ldots + \zeta_s = 0,$ where $\zeta_i^n = 1,$ then we can partition it in groups $\{\xi_1, \ldots, \xi_k\}$ of all $k$'th roots of some $z$ with $|z| = 1$ (with zero sum of elements in this group), with possibly different $k$ for different groups?
UPD
Example: if $\zeta_1 + \zeta_2 = 0,$ then $\zeta_1^2 = \zeta_2^2 = z,$ so $\zeta_1, \zeta_2$ are square roots of this $z.$ Is it not hard to show that if $\zeta_1 + \zeta_2 + \zeta_3 = 0,$ then $\zeta_1^3 = \zeta_2^3 = \zeta_3^3 = z,$ so $\zeta_1, \zeta_2, \zeta_3$ are cubic roots of some $z$.
If $\zeta_1 + \zeta_2 + \zeta_3 + \zeta_4 = 0,$ it seems there are two possiblities - like $\{1, i, -1, -i\}$ (one part), or like $\{e^{\pi i/3}, e^{4\pi i/3}\}, \{e^{2\pi i/3},e^{10\pi i/3}\}.$ But I don't know what is the argumentation for the general case.