Sum of Cube Roots of Unity

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Theorem

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

Then:

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


Proof 1

\(\ds 1 + \omega + \omega^2\) \(=\) \(\ds 1 + \paren {-\dfrac 1 2 + \dfrac {\sqrt 3} 2} + \paren {-\dfrac 1 2 - \dfrac {\sqrt 3} 2}\) Cube Roots of Unity
\(\ds \) \(=\) \(\ds 1 - \frac 1 2 - \frac 1 2 + \dfrac {\sqrt 3} 2 - \dfrac {\sqrt 3} 2\)
\(\ds \) \(=\) \(\ds 0\)

$\blacksquare$


Proof 2

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.

$\blacksquare$


Proof 3

Observe:

\(\ds \paren {1 - \omega} \paren {1 + \omega + \omega^2}\) \(=\) \(\ds 1 - \omega^3\) Difference of Two Cubes
\(\ds \) \(=\) \(\ds 1 - 1\) Cube Roots of Unity
\(\ds \) \(=\) \(\ds 0\)

As $\omega \ne 1$, it follows:

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

$\blacksquare$

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