Complex number problem with some behaviour I thought was interesting

Nice problem!

Let's think about the largest $q$ such that $z^q=1$ has no roots in the region $\frac{\sqrt2}{2} \le \Re{(z)} \le \frac{\sqrt3}{2}$ . Since all complex roots of unity satisfy $|z|=1$ , this restriction is equivalent to there being no roots with argument between $\frac{\pi}{6}$ and $\frac{\pi}{4}$ . (We only need to consider the first quadrant since the roots occur in conjugate pairs - ie they are symmetric about the real axis.)

The $q^{th}$ roots of unity have arguments $0,\frac{2\pi}{q},\frac{4\pi}{q},\ldots,\frac{2\pi k}{q},\ldots,\frac{2\pi (q-1)}{q}$ . We require that $\frac{\pi}{6} \le \frac{2\pi k}{q} \le \frac{\pi}{4}$ has no solutions, or in other words that $2q \le 24k \le 3q$ has no solutions for integer $k,q$ .

The largest integer $k$ such that $24k < 2q$ is $\left \lfloor \frac{q}{12} \right \rfloor$ . The smallest $k$ such that $24k > 3q$ is $\left \lceil \frac{q}{8} \right \rceil = \left \lfloor \frac{q}{8} \right \rfloor + 1$ (note that we can change to the floor function like this since if $8|q$ , $k=\frac{q}{8}$ gives a root in the restricted region)

We need there to be no integers between $\left \lfloor \frac{q}{12} \right \rfloor$ and $\left \lfloor \frac{q}{8} \right \rfloor + 1$ .

Since $\left \lfloor \frac{q}{12} \right \rfloor \le \left \lfloor \frac{q}{8} \right \rfloor$ , this can only happen if $\left \lfloor \frac{q}{12} \right \rfloor = \left \lfloor \frac{q}{8} \right \rfloor$ . The largest such $q$ is $15$ :

We have $\left \lfloor \frac{15}{12} \right \rfloor = \left \lfloor \frac{15}{8} \right \rfloor = 1$

For $16\le q \le 23$ , $\left \lfloor \frac{q}{12} \right \rfloor = 1$ and $\left \lfloor \frac{q}{8} \right \rfloor = 2$ .

For $q \ge 24$ , write $q=24a+b$ , where $a \ge 1$ and $0 \le b < 24$ . Then $\left \lfloor \frac{q}{12} \right \rfloor = 2a + \left \lfloor \frac{b}{12} \right \rfloor$ and $\left \lfloor \frac{q}{8} \right \rfloor = 3a + \left \lfloor \frac{b}{8} \right \rfloor$ , and

$\left \lfloor \frac{q}{8} \right \rfloor - \left \lfloor \frac{q}{12} \right \rfloor = 3a + \left \lfloor \frac{b}{8} \right \rfloor - 2a - \left \lfloor \frac{b}{12} \right \rfloor = a + \left \lfloor \frac{b}{8} \right \rfloor - \left \lfloor \frac{b}{12} \right \rfloor \ge a \ge 1$ (again using the fact that $\left \lfloor \frac{b}{12} \right \rfloor \le \left \lfloor \frac{b}{8} \right \rfloor$ ).

Returning to the problem, we have shown that $15$ is the largest $q$ such that no root of $z^q=1$ lies in the restricted region; the required answer is therefore $\boxed{p=16}$ .

Note that $|z|=1$ for almost all $n$ . Then $\text{Arg}(z)\in\pm\left[\frac{\pi}6,\frac{\pi}4\right]$ . Let $k\in\mathbb Z$ . We need
$\begin{aligned} n\times\text{Arg}(z)&=2\pi k\\ n&\in[8k,12k]\\ p&=\min(\{r:\forall(n\ge r)(\exists k\in\mathbb Z)(n\in[8k,12k])\})\\ &=16 \end{aligned}$

When dealing with complex numbers, it is often helpful to put the problem into a geometric context. The complex numbers $z$ such that $z^n = 1$ form a regular n-gon in the complex plane with one vertex at $z = 1$ . Therefore, in a geometric context, we are asked to find the smallest n-gon such that both it and all larger n-gons always have at least one vertex between $\mathbb{R}(z) = \frac{\sqrt{2}}{2}$ and $\mathbb{R}(z) = \frac{\sqrt{3}}{2}$ . (Note that this is not the same as asking for the smallest n-gon that has a vertex between $\mathbb{R}(z) = \frac{\sqrt{2}}{2}$ and $\mathbb{R}(z) = \frac{\sqrt{3}}{2}$ !) If we draw a line with length $1$ to $\mathbb{R}(z) = \frac{\sqrt{2}}{2}$ , we make an angle of $\frac{\pi}{4}$ with the real axis, and if we draw a line with length $1$ to $\mathbb{R}(z) = \frac{\sqrt{3}}{2}$ , we make an angle of $\frac{\pi}{6}$ with the real axis. Let $\theta$ be the central angle of our n-gon. We have that for some $k \in \mathbb{Z}$ , $\frac{\pi}{6} \le k\theta \le \frac{\pi}{4}$ . If we let $k = 1$ , we just have $\frac{\pi}{6} \le \theta \le \frac{\pi}{4}$ . Letting $k = 2$ , we have $\frac{\pi}{12} \le \theta \le \frac{\pi}{8}$ and with $k = 3$ we have $\frac{\pi}{18} \le \theta \le \frac{\pi}{12}$ . Note that $\frac{\pi}{12}$ occurs in both the inequality for $k=2$ and the inequality for $k=3$ ! This means that for all n-gons whose central angle $\theta$ is within the larger interval $\left[\frac{\pi}{18}, \frac{\pi}{8}\right]$ , at least one of their vertices lies between $\mathbb{R}(z) = \frac{\sqrt{2}}{2}$ and $\mathbb{R}(z) = \frac{\sqrt{3}}{2}$ . It is easy to see that as we continue with larger values of $k$ , the intervals will just continue overlapping. Therefore, the largest central angle possible is $\frac{\pi}{8}$ , and thus the smallest n-gon possible is a $16$ -gon. The answer is therefore $\boxed{p = 16}$ .