Properties of Cube Roots of Unity

Three identities memorised together. ω³ = 1 is the defining relation. 1+ω+ω² = 0 follows from the sum of cube roots of unity (cn32 with n=3) — it is the most-used identity in JEE. ω² = ω̄ since ω and ω² are conjugates.

Two identities. Every problem involving cube roots of unity reduces to one of them.

Identity 1 — $\omega^3 = 1$

$\omega$ is defined as a root of $z^3 = 1$. So by definition:

$$\omega^3 = 1$$

This means any power of $\omega$ reduces to $\omega^0$, $\omega^1$, or $\omega^2$ by dividing the exponent by 3 and taking the remainder:

$$\omega^{3k} = 1, \quad \omega^{3k+1} = \omega, \quad \omega^{3k+2} = \omega^2$$

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

The three cube roots of unity are the roots of $z^3 - 1 = 0$. Factor:

$$z^3 - 1 = (z - 1)(z^2 + z + 1)$$

Since $\omega \neq 1$, it is a root of the quadratic factor:

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

Alternatively: this is the $n = 3$ case of the general result that the sum of all $n$th roots of unity is zero (cn32). $\blacksquare$

Identity 3 — $\omega^2 = \bar{\omega}$

The two complex cube roots are:

$$\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \qquad \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$$

They are conjugates of each other: $\omega^2 = \bar{\omega}$.

This also follows from $\omega^2 = 1/\omega$ (since $\omega^3 = 1$), combined with $1/\omega = \bar{\omega}/|\omega|^2 = \bar{\omega}$ (since $|\omega| = 1$).

How to Use These in Problems

The two identities generate all other results:

Replacing $\omega^2$: From $1 + \omega + \omega^2 = 0$, write $\omega^2 = -1 - \omega$. Substitute wherever $\omega^2$ appears to reduce expressions to linear form in $\omega$.

Factoring: $a^3 - b^3 = (a-b)(a-\omega b)(a - \omega^2 b)$ — the three linear factors over $\mathbb{C}$ corresponding to the three cube roots of $b^3$.

Evaluating at $\omega$: If $f(x) = a_0 + a_1 x + a_2 x^2 + \cdots$, then $f(1) + f(\omega) + f(\omega^2)$ equals $3$ times the sum of coefficients of $x^{3k}$ terms. This is the selective-sum technique.