Cyclic Group of the Nth Roots of Unity

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I'm still reading Needham's book Visual Complex Analysis. There is a part about the geometric interpretation of the nth 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 nth 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.