I'm still reading Needham's book Visual Complex Analysis.
There is a part about the geometric interpretation of the *n*th roots of unity, that is, the solutions to the following equation.
$$
z^n = 1
$$

Using Euler's formula, we can represent the complex number *z* as \(r e^{i\theta}.\)
Taken to the power of a natural number *n*, we get \(r^n e^{i n \theta}.\)
To satisfy the equation, \(r^n\) must be one, so all solutions *z* must lie on the unit circle.
Multiplication of any complex number on the unit circle by itself rotates it on the circle by its angle.
The solutions *z* have the special property that a product of *n* identical factors *z* ends up on point \(1 = e^{i m \cdot 2 \pi}\) (*m* being any integer).
In other words: a point \(z = e^{i \theta}\) on the unit circle is a solution if \(n \theta\) is an integer multiple of \(2 \pi.\)
$$
\theta_m = \frac{m \cdot 2\pi}{n}
$$

Each integer *m* gives us an angle \(\theta_m\) such that \(e^{i\theta_m}\) is a solution *z*, but only the solutions arising from \(m = 0, 1, \ldots, n-1 \) are unique because \(e^{i (\theta_m - \theta_n)} = e^{i \theta_m}.\)
So the solutions are *n* equally spaced points on the unit circle with the first one always being \(e^0 = 1.\)
Connecting adjacent points gives us a regular polygon.

This reminded me of a bit of abstract algebra I came across some time ago.
The *n*th roots of unity form a cyclic group.
A generator of this group is \(a = e^{i \frac{2\pi}{n}}\) and the least positive integer *m* such that \(a^m\) is the identity element \(1\) is \( m = n,\) so the group is of order *n*.

Needham, T. (1997). *Visual complex analysis*. Oxford University Press.