Sum of roots of unity - Alex's public notes

Lemma

For any $n$th root of unity $\omega$ where $\omega \not = 1$ we have

$$ \sum_{i=0}^{n-1} \omega^i = 0.$$

Proof

For any number $z$ we have

$$(z - 1)(1 + z + z^2 + \ldots + z^{n-1}) = z^n - 1.$$

Therefore

$$ (\omega - 1) \cdot \left ( \sum_{i=0}^{n-1} \omega^i \right ) = \omega^n - 1 = 0$$

however as $\omega \not = 1$ we have $\omega - 1 \not = 0$ giving the desired result.