Sum of Cube Roots of Unity/Proof 1

Theorem

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

Then:

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

Proof

\(\ds 1 + \omega + \omega^2\)

\(=\)

\(\ds 1 + \paren {-\dfrac 1 2 + \dfrac {\sqrt 3} 2} + \paren {-\dfrac 1 2 - \dfrac {\sqrt 3} 2}\)

Cube Roots of Unity

\(\ds \)

\(=\)

\(\ds 1 - \frac 1 2 - \frac 1 2 + \dfrac {\sqrt 3} 2 - \dfrac {\sqrt 3} 2\)

\(\ds \)

\(=\)

\(\ds 0\)

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 3$. Roots of Unity