If $\ z_k\$ are complex numbers, sort of uniformly spread out on the unit circle, then what is $\ \sup \{\ \vert z_0 + \ldots + z_{n-1} \vert\ \}\ ?$

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.

• My numerical simulations suggest that $s_n=2$ for all $n$, but so far I have only been able to prove a weaker bound. Jan 23 at 16:23

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.

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 $$(*)$$.