Proving equations involving the powers of a complex cube root of unity ω

Question

The question in this homework problem is to show

$ω^4 + ω^5 = -ω^6$

given that $ω$ is a complex cube root of unity.

I am also required to show that $(1 - ω)^2 = -3ω$, but if I am assisted with the approach to take with this kind of problem I think I could manage on my own.

I know the complex cube roots of unity to be $\frac{-1}{2} ± \frac{\sqrt{3}}{2}i$, but I feel as if picking one and plugging it in instead of $ω$ would be an unnecessarily complicated way of proving the equations given.

I also know that if $ω$ is a complex cube root of unity, $ω^2$ is the other, i.e. the complex conjugate of $w$, and that the sum of the three cube roots of unity is 0.

Answer 1

From Vieta's formula, $\omega_1+\omega_2+1 = 0$, where $\omega_1,\omega_2$ are two distinct complex roots of unity.

You already knew $\omega_1 = \omega_2^2$ and $\omega_2=\omega_1^2$. Either way you get $\omega+\omega^2+1 = 0$, and multiply both sides by $\omega^4$ to get the result.

Answer 2

As $\displaystyle x^3=1\implies (x-1)(x^2+x+1)=0$ and $\displaystyle \omega\ne1 $

$\displaystyle \omega,\omega^2$ are the roots of $x^2+x+1=0$

Applying Vieta's formula to find $\displaystyle \omega+\omega^2$ and $\omega\cdot \omega^2=1$

Now, $\displaystyle \omega^4+\omega^5=\omega\cdot1+\omega^2\cdot1=\cdots$