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Problem:

Let $z_1, z_2, ..., z_n$ be complex numbers with a modulus of $1$.

Given the condition $S = \sum_{j=1}^{n} z_j = 0$. Please compute all possible values for $z_j$.

My idea:

Assume $$z_j = e^{i\theta_j},\quad \theta_j \in [0,2\pi),\quad j=1,...,n$$ It can be calculated that $$|S|^2 = \left( \sum_{j=1}^{n}\cos\theta_j \right)^2 + \left( \sum_{j=1}^{n}\sin\theta_j \right)^2 = n + \sum_{1 \le j < k \le n} 2 \cos({\theta_j - \theta_k})$$ According to this formula, when $n=2$, we have $2 + 2\cos(\theta_1 - \theta_2) = 0$, so $|\theta_1 - \theta_2| = \pi$. This completes the $n=2$ case.

When n is larger, I don't know how to handle it.

I want to handle arbitrary $n$ cases. Thanks for any suggestions. Recommendation on relevant literature is also appreciated.

sunset
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  • Hi, welcome to Math SE. For $n\ge3$, this problem is equivalent to the following: find all (possibly degenerate) planar $n$-gons with all sides of length $1$. The existence of rhombuses suggests this isn't trivial. – J.G. Feb 29 '24 at 07:18
  • There will be a lot of solutions. Note that a rotation of a solution will also be a solution. So, you could assume that $z_1 = 1$. With this assumption, the $n = 3$ becomes easier: the cube roots of $1$. $n = 4$ raises a different problem, it could be two pairs of $n = 2$ solutions. So, you might like to add the condition that no subset sums to $0$. – badjohn Feb 29 '24 at 07:24
  • @J.G. Thanks for your comment. Any further advice on finding planar polygons with all sides of length 1? – sunset Feb 29 '24 at 07:40
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    I don't think there's a helpful complete characterization of them. Even the $n=5$ case is diverse. – J.G. Feb 29 '24 at 09:22
  • The sum can be composed of sub-sums yielding zero, any sub-sum being invariant wrt to its own 2d-rotation group. So this starts a program to characterize n-dimensional representations of the rotation group – Roland F Feb 29 '24 at 09:54

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