Cube Roots of Unity

Cube roots of unity

Number $1$ is often called unity. Let $x$ be a cube root of unity, i.e. $x^3=1$. Then:

From $x-1=0$: $x=1$. From $x^2+x+1=0$, using the quadratic formula: $x=\dfrac{-1\pm\sqrt{1-4}}{2}=\dfrac{-1\pm\sqrt{-3}}{2}=\dfrac{-1\pm\sqrt3\,i}{2}$.

So the cube roots of unity are

Among these three, one is real ($1$) and the other two are complex conjugates of each other. It can be verified directly (by squaring) that $\left(-\dfrac12+\dfrac{\sqrt3}{2}i\right)^2=-\dfrac12-\dfrac{\sqrt3}{2}i$, and similarly the square of the second root gives back the first — so the two complex roots square into each other.

Let $\omega=-\dfrac12+\dfrac{\sqrt3}{2}i$; then $\left(-\dfrac12-\dfrac{\sqrt3}{2}i\right)=\omega^2$. So the cube roots of unity are $1,\omega,\omega^2$, where

Also note that $1=e^{2\pi i}$, $\omega=e^{2\pi i/3}$, $\omega^2=e^{4\pi i/3}$ — connecting this section directly to the exponential form of Section 1.5.5.

Properties of $1,\omega,\omega^2$ — the standing toolkit used throughout Exercise 1.4 and the Miscellaneous Exercise:

1. $\omega$ is a complex cube root of $1$, so $\omega^3=1$.
2. From $\omega^3-1=0$, factoring gives $(\omega-1)(\omega^2+\omega+1)=0$; since $\omega\neq1$, it must be that $\omega^2+\omega+1=0$. (Equivalently, $1+\omega+\omega^2=0$ — the single most-used identity of the section.)
3. $\omega^2=\dfrac{1}{\omega}$ and $\dfrac{1}{\omega^2}=\omega$ — the two nonreal cube roots are each other's reciprocals as well as each other's conjugates.
4. $\omega^3=1$, so $\omega^{3n}=1$ for any integer $n$ — powers of $\omega$ cycle every $3$ steps, exactly as powers of $i$ cycle every $4$.
5. $\omega^4=\omega^3\cdot\omega=\omega$, so $\omega^{3n+1}=\omega$.
6. $\omega^5=\omega^2\cdot\omega^3=\omega^2\cdot1=\omega^2$, so $\omega^{3n+2}=\omega^2$.
7. $\bar\omega=\omega^2$ (the conjugate of $\omega$ is exactly $\omega^2$, since they are complex conjugates by construction).
8. $\overline{\omega^2}=\omega$ (the conjugate relationship works both ways). Worked Example 1: if $\omega$ is a complex cube root of unity, prove that (i) $\dfrac1\omega+\dfrac{1}{\omega^2}=-1$ (ii) $(1+\omega^2)^3=-1$ (iii) $(1-\omega+\omega^2)^3=-8$. Since $\omega^3=1$ and $\omega^2+\omega+1=0$: $\omega^2+1=-\omega$ and $\omega+1=-\omega^2$.

(i) $\dfrac1\omega+\dfrac{1}{\omega^2}=\dfrac{\omega+1}{\omega^2}=\dfrac{-\omega^2}{\omega^2}=-1$.

(ii) $(1+\omega^2)^3=(-\omega)^3=-\omega^3=-1$.

(iii) $(1-\omega+\omega^2)^3=(1+\omega^2-\omega)^3=(-\omega-\omega)^3$ (using $1+\omega^2=-\omega$) $=(-2\omega)^3=-8\omega^3=-8\times1=-8$.

Worked Example 2: if $\omega$ is a complex cube root of unity, show that (i) $(1-\omega+\omega^2)^5+(1+\omega-\omega^2)^5=32$ (ii) $(1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^5)=9$.

(i) $(1-\omega+\omega^2)^5=(-\omega-\omega)^5=(-2\omega)^5=-32\omega^5$. $(1+\omega-\omega^2)^5=(-\omega^2-\omega^2)^5$ (using $1-\omega^2=...$ actually $\omega+1=-\omega^2$, so $1+\omega-\omega^2=-\omega^2-\omega^2=-2\omega^2$) $=(-2\omega^2)^5=-32\omega^{10}$. Sum: $-32\omega^5-32\omega^{10}=-32\omega^5(1+\omega^5)=-32\omega^5(1+\omega^2)$ (since $\omega^5=\omega^2$) $=-32\omega^5\times(-\omega)=32\omega^6=32(\omega^3)^2=32(1)^2=32$.

(ii) $(1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^5)=(1-\omega)(1-\omega^2)(1-\omega^3\omega)(1-\omega^3\omega^2)=(1-\omega)(1-\omega^2)(1-\omega)(1-\omega^2)=(1-\omega)^2(1-\omega^2)^2=[(1-\omega)(1-\omega^2)]^2$. Now $(1-\omega)(1-\omega^2)=1-\omega^2-\omega+\omega^3=1-(\omega^2+\omega)+1=1-(-1)+1=3$. So the whole product is $3^2=9$.

Worked Example 3: if $\omega$ is a complex cube root of unity such that $x=a+b,\ y=a\omega+b\omega^2,\ z=a\omega^2+b\omega$ ($a,b\in\mathbb{R}$), prove (i) $x+y+z=0$ (ii) $x^3+y^3+z^3=3(a^3+b^3)$.

(i) $x+y+z=a+b+a\omega+b\omega^2+a\omega^2+b\omega=a(1+\omega+\omega^2)+b(1+\omega+\omega^2)=a(0)+b(0)=0$. …