Cube Roots of Unity and Their Properties

The $n=3$ case of the previous section deserves its own name because it shows up so often in algebraic identities: the cube roots of unity. Taking $\omega = \cos\dfrac{2\pi}{3}+i\sin\dfrac{2\pi}{3}$ and evaluating the cosine and sine at $120°$ gives

so the three cube roots of unity are $1,\ \omega,\ \omega^2$. Notice $\omega^2$ is just the complex conjugate of $\omega$, which makes sense — the three roots must be symmetric about the real axis since $1$ is real.

A handful of algebraic facts about $\omega$ are used again and again, and it is worth internalising them rather than re-deriving them every time:

• $\omega^3=1$ (that's the whole point of $\omega$ being a cube root of unity), and more generally $\omega^{3m}=1,\ \omega^{3m+1}=\omega,\ \omega^{3m+2}=\omega^2$ for any integer $m$ — powers of $\omega$ simply cycle through $\{1,\omega,\omega^2\}$ with period 3.
• $1+\omega+\omega^2=0$. This is just the $n=3$ case of the 'sum of roots of unity is zero' fact from §2.2, but it is so heavily used in simplification that it is worth remembering on its own — it is the standard trick for eliminating $\omega^2$ from an expression (replace $\omega^2$ by $-1-\omega$) or vice versa.
• Geometrically, $1,\omega,\omega^2$ are the three vertices of an equilateral triangle inscribed in the unit circle.
• The cube roots of any positive real number $a$ are $a^{1/3},\ a^{1/3}\omega,\ a^{1/3}\omega^2$ — take the ordinary positive real cube root and multiply it by each cube root of unity in turn.

Worked example. If $1,\omega,\omega^2$ are the cube roots of unity, show that $(2-\omega)(2-\omega^2)=7$ and use it to simplify $(1-\omega+\omega^2)(1-\omega^2+\omega)$.

For the first part, expand directly: $(2-\omega)(2-\omega^2) = 4 - 2\omega^2-2\omega+\omega^3 = 4-2(\omega+\omega^2)+\omega^3$. Now substitute $\omega+\omega^2=-1$ (from $1+\omega+\omega^2=0$) and $\omega^3=1$: this gives $4-2(-1)+1 = 4+2+1=7$, as required. …