# Confinement result for unit complex numbers

Inspired by this problem, and some computer simulations, I almost convinced myself of the following result. However, I am coming short on a proof.

Result: Let $$n\geq 3$$, and $$2n+1$$ complex numbers $$z_{1},\ldots ,z_{2n+1}$$ on the unit circle such that $$\sum_{k=1}^{2n+1} z_{k} =0$$.

There exists $$\mathscr{I} \subset \{ 1,\ldots, 2n+1 \}$$, such that $$|\mathscr{I}| = n$$ and $$\left| \sum_{i \in \mathscr{I}} z_{i}\right| \leq \frac{n-1}{n}.$$

1. Is this result true?
2. Is it known?
3. If so, any proof?

Note: There are other confinement results for instance the Polygonal Confinement Theorem.

• What if $n=2$ and $z_1, \ldots, z_5$ are the fifth root of unity? I don't think that $|z_j+z_k| \le 1/2$ for any two of them. Sep 18, 2023 at 17:25
• If you define $C_n$ as the smallest possible value of $\left| \sum_{i \in \mathscr{I}} z_{i}\right|$ then I my guess would be that $C_n$ decreases with $n$. Sep 18, 2023 at 17:58
• @MartinR: maybe, but we already know that $C_n \leq 1$, so… Sep 18, 2023 at 18:11
• @MartinR I changed $n\geq 3$ instead of $n\geq 2$. I think $C_n$ increases with $n$. Sep 19, 2023 at 3:24
• Can you share your code for the computer simulation? Sep 20, 2023 at 12:05

Let $$n=9$$, $$\xi$$ be a nonreal cubic root of $$1$$ and $$z_k=\xi^k$$ for each $$k\in\{1,\dots,9\}$$. It is easy to check that for each $$\mathscr{I} \subset \{ 1,\ldots, 9 \}$$ with $$|\mathscr{I}| = 4$$ we have $$\left| \sum_{i \in \mathscr{I}} z_{i}\right|\ge 1.$$
• Thanks Alex, so the upper bound $\frac{n-1}{n}$ in the statement above is wrong. What about the upper bound 1 instead (which was from the initial problem linked above)? Nov 8, 2023 at 1:08