The nth roots of unity

Statement

Let $n$ be a positive integer. A complex number $z$ is an $n$th root of unity if and only if $$z = e^{2 \pi i k / n} = \cos \frac{2 \pi k}{n} + i \sin \frac{2 \pi k}{n}$$ for some integer $k$ with $0 \le k < n$.

Remarks

Proof: by de Moivre's formula, each such $z$ satisfies $z^n = e^{2 \pi i k} = 1$. Conversely, $z^n = 1$ forces $|z| = 1$ and $n \arg z \in 2 \pi \mathbb{Z}$. For $n = 3$, the roots are $1$ and $-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$.

Explanations

L1 human

The $n$th roots of unity sit on the unit circle at equal angles of $2\pi/n$, starting at $1$. For $n = 6$, they are the six corners of a regular hexagon.