Maybe a union of odd regular polygons ?

Working along the unit circle, any set of points $S = \{e^{2\pi i a_k}\}_{k=1,2,...,n}$ works if for some suitable choice of signs $\epsilon_k\in \{-1,+1\}$ we have $\sum_{k=1}^n \epsilon_k e^{\pi i a_k} = 0$. The equivalence is not too hard to demonstrate, but I'm not sure whether there's a better interpretation of this condition.

DVDthe1st wrote: Working along the unit circle, any set of points $S = \{e^{2\pi i a_k}\}_{k=1,2,...,n}$ works if for some suitable choice of signs $\epsilon_k\in \{-1,+1\}$ we have $\sum_{k=1}^n \epsilon_k e^{\pi i a_k} = 0$. The equivalence is not too hard to demonstrate, but I'm not sure whether there's a better interpretation of this condition. Well, but the problem isn't equivalent with this actually... Well, Yeah, that's easy to get... And I thought that with rotation, we should get that condition too. So, I thought that it should be union of odd regular polygons,,, Well, maybe no Edit : I'm sorry you were right Well, I think there won't be a better interpretation for that

anyone has full solution ? thank you