Cube Roots of Unity

$\omega=\dfrac{-1+i\sqrt3}{2},\;\omega^2=\dfrac{-1-i\sqrt3}{2}=\bar\omega$. The two identities:

• $\omega^3=1$ — so $\omega^n=\omega^{\,n\bmod 3}$ (powers cycle every 3).
• $1+\omega+\omega^2=0$ — so $\omega+\omega^2=-1$ and $\omega^2=-1-\omega$.

Also $\omega\cdot\omega^2=1$ (they are reciprocals/conjugates), and $|\omega|=1$. The $-2\omega$ collapses (one step from $1+\omega+\omega^2=0$, and the shape of every 'raise to the 100th power' question): $1+\omega-\omega^2=-2\omega^2$, $1-\omega+\omega^2=-2\omega$, $-1+\omega+\omega^2=-2$, $1+\omega=-\omega^2$, $1+\omega^2=-\omega$. So $(1+\omega-\omega^2)^{3n}=(-2\omega^2)^{3n}=(-2)^{3n}$, and $(1-\omega+\omega^2)^5+(1+\omega-\omega^2)^5=-32(\omega^5+\omega^{10})=-32(\omega^2+\omega)=32$. A symmetric-function identity: with $x=a+b$, $y=a\omega+b\omega^2$, $z=a\omega^2+b\omega$: $x+y+z=0$, $xyz=a^3+b^3$, $xy+yz+zx=-3ab$, and hence $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=6ab$.

• Reduce then collapse: $\omega^n=\omega^{\,n\bmod3}$, then apply $1+\omega+\omega^2=0$.
• Quadratic roots: $x^2+x+1=0\Rightarrow x=\omega,\omega^2$; $x^2-x+1=0\Rightarrow x=-\omega,-\omega^2$ (primitive 6th roots).
• Cube roots of $k$: the roots of $z^3=k$ are $k^{1/3},\,k^{1/3}\omega,\,k^{1/3}\omega^2$ — they sum to 0 and form an equilateral triangle.
• Sums like $\alpha^n+\beta^n$ for $\alpha,\beta$ cube/6th-roots are a small-case match on $n\bmod 3$ (or 6).
• The homogeneous form: $z_1^2+z_1z_2+z_2^2=0$ — divide by $z_2^2$ to get $t^2+t+1=0$ with $t=z_1/z_2$, so $z_1=\omega z_2$ or $\omega^2z_2$: the two numbers have equal modulus and arguments differing by $\tfrac{2\pi}3$ (with 0 they form an equilateral triangle). Likewise $z_1^2-z_1z_2+z_2^2=0$ gives $z_1=-\omega z_2$ or $-\omega^2 z_2$, i.e. $z_1/z_2=e^{\pm i\pi/3}$.
• Coefficients in GP: if $a,b,c$ are in GP ($b^2=ac$) the roots of $ax^2+bx+c=0$ satisfy $\dfrac{\alpha}{\beta}=\omega$ or $\omega^2$ — the discriminant is $b^2-4ac=-3ac$, and dividing through by $c$ with $x=\sqrt{c/a}\,t$ gives $t^2+t+1=0$.