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.
Best Answer
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.