“Converse” to the theorem “sum of roots of unity equal 0”

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.

• I'm confused. $\exp(i\pi/3)$ and $\exp(4i\pi/3)$ are both cube roots of unity. Are you trying to restrict to square roots? – YiFan Dec 21 '18 at 13:56
• By sum of roots of unity, of course it refers to the set of all roots of unity corresponding to some number $n$, so in this case I mean that $\pi/3$ and $4\pi/3$ belong to a complete set of roots of unity corresponding to 2, after a rotation. Of course it needn't only be square roots. – enochk. Dec 21 '18 at 15:02
• I see. Sorry, I misunderstood on my first reading. – YiFan Dec 21 '18 at 15:03

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.
• Thanks for the comment. First, if self-intersecting, rearrange until not self intersecting. Second, for $\exp(i\phi_1)+\exp(i\phi_2)$ to be a vertex, $\phi_2$ is uniquely defined by $\phi_1$ if vertices are ordered counterclockwise. Third, the vertices do not have to correspond to one complete set of roots of unity ( that is $\exp(2\pi i n /N)$ for some $N$) but they might be a complete set after a rotation, or the may be made up of several complete sets of roots of unity for different $N$'s. Fourth, the only cyclic rhombus is a square, which correspond to the roots of unity of N=4. – enochk. Dec 21 '18 at 23:17
• @enoch: I don't follow your last comment. There's no need for the polygon to be cyclic. Very explicitly, consider $1 + z + (-1) + (-z)$ for any $z = \exp(i \phi)$. – Qiaochu Yuan Dec 21 '18 at 23:22
• Your example is a cyclic polygon and not technically a rhombus, but I get your point. The polygon must be cyclic by definition because its vertices are on the circle. Also what you proposed is made of two sets $\{1,-1\}$ and $\{z, -z\}$. They are both roots of unity for $N=2$ only that the second set requires a rotation. You can see that I used the same example with $z=\exp(i \pi/3)$ in the main post. – enochk. Dec 22 '18 at 0:06
• @enochk. You seem to have misunderstood what polygon Qiaochu is referring to. When a sum $a_0+a_1+\dots+a_n$ is $0$, the intended polygon's vertices are not $a_1,a_2,\dots,a_n$ but rather $0, a_1,a_1+a_2,a_1+a_2+a_3,\dots,a_1+a_2+\dots+a_{n-1}$. So the example in his comment is indeed a rhombus and, for most values of $z$, not cyclic. – Andreas Blass Dec 22 '18 at 0:24
• Thanks for the clarification. However, the example given is still not a counter-example to my main query as the example is a sum of two complete sets of roots of unity for $N=2$; $z$ and $-z$ are simply a rotated set of square roots of unity. Expanding on Qiaochu's visualisation for a sum of 3 terms, if $\exp(i \phi_1)+\exp(i \phi_2) + \exp(i \phi_3)=0$, the partial sums must correspond to the vertices of an equilateral triangle, hence taking WLOG $\phi_1=0$, the other points must be $\zeta_3$ and $\zeta_3^2$. – enochk. Dec 22 '18 at 1:14