Sum of Cube Roots of Unity/Proof 2

Theorem

Let $U_3 = \set {1, \omega, \omega^2}$ denote the Cube Roots of Unity.

Then:

$1 + \omega + \omega^2 = 0$

Proof

From Sum of Powers of Primitive Complex Roots of Unity:

$\ds \sum_{j \mathop = 0}^{n - 1} \alpha^{j s} = \begin {cases} n & : n \divides s \\ 0 & : n \nmid s \end {cases}$

Here we have that $n = 3$ and $s = 1$.

Thus $n$ is not a divisor of $s$.

Hence the result.

$\blacksquare$