# Cyclic Group of the Nth Roots of Unity

| math

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.