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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Martin R
    Sep 18, 2023 at 17:25
  • $\begingroup$ 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 $n$. $\endgroup$
    – Martin R
    Sep 18, 2023 at 17:58
  • $\begingroup$ @MartinR: maybe, but we already know that $C_n \leq 1$, so… $\endgroup$
    – Aphelli
    Sep 18, 2023 at 18:11
  • $\begingroup$ @MartinR I changed $n\geq 3$ instead of $n\geq 2$. I think $C_n$ increases with $n$. $\endgroup$ Sep 19, 2023 at 3:24
  • $\begingroup$ Can you share your code for the computer simulation? $\endgroup$
    – Martin R
    Sep 20, 2023 at 12:05

1 Answer 1


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 $\left| \sum_{i \in \mathscr{I}} z_{i}\right|\ge 1.$

  • 1
    $\begingroup$ 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)? $\endgroup$ Nov 8, 2023 at 1:08
  • $\begingroup$ @NathanPortland The solution idea proposed there looks OK. $\endgroup$ Nov 10, 2023 at 3:49

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