Sum of the roots of unity

Statement

Let $n \ge 2$ be an integer and let $\zeta$ be a primitive root of unity. Then $$\sum_{k=0}^{n-1} \zeta^k = 1 + \zeta + \zeta^2 + \dots + \zeta^{n-1} = 0.$$

Remarks

Proof: $\zeta \neq 1$, so the geometric sum formula gives $\frac{\zeta^n - 1}{\zeta - 1} = 0$. Thus the sum of all the $n$th roots of unity is $0$ for $n \ge 2$. For $n = 1$ the sum is $1$.

Explanations

L1 human

The $n$th roots of unity are the corners of a regular polygon with center $0$. By symmetry, their center of mass is $0$, and so their sum is $0$. For $n = 4$: $1 + i + (-1) + (-i) = 0$.