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Let $\ n\in\mathbb{N}\ $ and suppose $\ z_k = x_k + iy_k,\ $ where $ \vert z_k \vert = 1\ $ and $\ \frac{2k\pi}{n} < \arg(z_k) < \frac{2(k+1)\pi}{n}\quad $ for all $\ k\in \{ 0,\ \ldots,\ n-1 \}.$

What is $\ s_n = \sup \{\ \vert z_0 + \ldots + z_{n-1} \vert\ \}\ $ in terms of $\ n\ ?$

Also, what is $\ \lim _{n\to\infty} s_n\ ?$

What I do know is that $\ n\ $ uniformly spread out points around a circle have the property: $\ z_1+\ldots+z_n = 0.$

At first, I just thought we put all points as far right as possible. But I don't think this is correct for all $\ n.\ $ For example, for $\ n=16,\ $ if we move all points as far to the right as possible, then I think we can make $\ \vert z_0 + \ldots + z_{n-1} \vert\ $ larger by moving $\ z_8\ $ clockwise towards the negative real axis, although maybe I am wrong about this.

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  • $\begingroup$ My numerical simulations suggest that $s_n=2$ for all $n$, but so far I have only been able to prove a weaker bound. $\endgroup$
    – Martin R
    Jan 23 at 16:23
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Here is a way to proceed, with quite a lot still to fill in.

Suppose that you have a configuration close to the maximum modulus and consider $S$ the sum and one component of the sum $z$ so that $S=z+(S-z)$. Now consider $S-z$ as fixed and maximise the modulus of $S$ by changing $z$. Considering arguments, I think you will find that you need $z$ to point as close to the direction of $S-z$ as possible.

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I can show that $$ \tag{$*$} 2 \le s_n \le \frac{2+ \sin(\pi/(2n))}{\cos(\pi/n)} $$ for all $n$, and that implies $\lim _{n\to\infty} s_n = 2$.

My conjecture is that $s_n = 2$, but I haven't yet been able to prove it.

For continuity reasons, we can relax the conditions on $z_n$ to $$ \frac{2k\pi}{n} \le \arg(z_k) \le \frac{2(k+1)\pi}{n}\, . $$ That does not change the supremum (and in fact makes it a maximum).

Let $\omega = e^{2\pi i/n}$ be the primitive $n$-th root of unity. For the lower bound, we pick the $z_k$ with minimal real part. The choice $$ z_1 + \cdots + z_n = \omega^1 + \cdots + \omega^{k} + \omega^{k} + \cdots + \omega^{n-1} = -2 $$ for even $n = 2k$, and $$ z_1 + \cdots + z_n = \omega^1 + \cdots + \omega^{k} + (-1) + \omega^{k+1} + \cdots + \omega^{n-1} = -2 $$ for odd $n=2k+1$ shows that $s_n \ge 2$.

For the upper bound let $z_1, \ldots, z_n$ be any admissible choice. Multiplying all points with $\omega$ gives another admissible choice. Therefore we can assume that the argument of $z_1 + \cdots + z_n$ is in the range $[-\pi/n, \pi/n]$. This allows us to get an estimate for the absolute value of the sum from an estimate for the real part of the sum. We have $$ \operatorname{Re}(z_1 + \ldots z_n) \le \operatorname{Re}(\omega^0 + \cdots + \omega^{k-1} + \omega^{k+1} + \cdots + \omega^{n}) = 2 $$ for even $n=2k$, and $$ \operatorname{Re}(z_1 + \ldots z_n) \le \operatorname{Re}(\omega^0 + \cdots + \omega^{k} + \omega^{k+2} + \cdots + \omega^{n}) \\ = 1 - \cos\left( \frac{(k+1)\pi}{2n}\right) = 2 + \sin\left( \frac{\pi}{2n}\right) $$ for odd $n=2k+1$. In any case, $$ 2 + \sin\left( \frac{\pi}{2n}\right) \ge \operatorname{Re}(z_1 + \ldots z_n) \ge \cos\left(\frac{\pi}{n}\right) |z_1 + \cdots + z_n| $$ and that proves the upper bound in $(*)$.

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