It is well known that sum of roots of unity equal 0. However, if $\sum_j \exp(i \phi_j)=0$, can we say something about the relation between the $\exp(i \phi_j)$'s? For example, we can rotate one of them to the position of 1, can we say that the other units, at least a subgroup of them, fall on the roots of unity?
Suppose we take the angles $0, \pi/3, \pi, 4 \pi/3$. Then $\exp(0) + \exp(i \pi/3) + \exp(i \pi) + \exp(i 4\pi/3) = 0$, but we can separate these units into the sets $\{0, \pi\}$ and $\{\pi/3, 4\pi/3\}$, the former of which correspond to roots of unity and the latter correspond to roots of unity after rotation.
My feeling is that there is an obvious counter-example, but so far I haven't found it. Any thoughts are appreciated.
Geometrically such a sum corresponds to a polygon (not necessarily convex, possibly self-intersecting) all of whose sides have length $1$; the vertices of the polygon are $0, \exp(i \phi_1), \exp(i \phi_1) + \exp(i \phi_2)$, etc. The angles involved need not be roots of unity; for example it could be a rhombus. The case of all of the roots of unity summing to zero corresponds to a regular $n$-gon.