Q.If $\omega\ne1$ is a cube root of unity, show that $\dfrac{a+b\omega+c\omega^2}{b+c\omega+a\omega^2}+\dfrac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2}=-1$.

$n$th roots of a complex number. If $\omega=\rho(\cos\phi+i\sin\phi)$ satisfies $\omega^n=z$ where $z=r(\cos\theta+i\sin\theta)$, write $z$ with its general argument $\theta+2k\pi$ (since a single value of $\theta$ would miss roots) and apply de Moivre's theorem to $\omega^n$: comparing modulus and argument gives $\rho^n=r$ and $n\phi=\theta+2k\pi$, so

Hence the $n$th roots of $z=r(\cos\theta+i\sin\theta)$ are

Although $k$ can be any integer, only $k=0,1,\dots,n-1$ give distinct values (larger $k$ repeats the same $n$ roots cyclically). Geometrically, all $n$ roots share the modulus $r^{1/n}$, so they lie on a circle of radius $r^{1/n}$ centred at the origin, equally spaced at angular intervals of $\dfrac{2\pi}n$ — i.e. at the vertices of a regular $n$-gon.

$n$th roots of unity. Setting $z=1=\cos0+i\sin0$ specialises the formula to

Writing $\omega=e^{2\pi i/n}$ (the primitive $n$th root), the $n$ roots are exactly $1,\omega,\omega^2,\dots,\omega^{n-1}$ — a geometric progression with common ratio $\omega$ — and they are the vertices of a regular $n$-gon inscribed in the unit circle.

Standing facts about the $n$th roots of unity (all provable from the GP sum/product formulas):

• Sum $1+\omega+\omega^2+\cdots+\omega^{n-1}=0$ (a finite GP with ratio $\omega\ne1$, sum $\dfrac{\omega^n-1}{\omega-1}=\dfrac{1-1}{\omega-1}=0$).
• Product $1\cdot\omega\cdot\omega^2\cdots\omega^{n-1}=(-1)^{n-1}$.
• They all satisfy $|z|=1$ and $z^n=1$.

Cube roots of unity ($n=3$). $1,\ \omega=\dfrac{-1+i\sqrt3}2,\ \omega^2=\dfrac{-1-i\sqrt3}2$, with the identities $1+\omega+\omega^2=0$ and $\omega^3=1$ used constantly to simplify expressions in $\omega$ (e.g. $\omega^4=\omega,\ 1+\omega=-\omega^2$).

Solving a binomial/related equation. Equations like $z^n=c$ (e.g. $z^3+27=0\iff z^3=-27$) are solved by writing $c$ in polar form and applying the root formula directly. A shifted equation like $(z-1)^3+8=0$ is solved by substituting $w=z-1$, solving $w^3=-8$ as $w=-2\times(\text{a cube root of unity})$, then recovering $z=1+w$ — this is why such roots naturally come out expressed in terms of $\omega$.