I have to do such a question:
Let ABC be a triangle, whose vertices A, B, C correspond to the complex numbers α, β, γ (the origin is not necessarily at one of the vertices), respectively. Let $ ω = e^{2πi/3 } $.
(a) Prove that $ α+ωβ +ω^2γ = 0 $ or $ α+ω^2β +ωγ = 0 $ if △ABC is equilateral.
(b) Prove that △ABC is equilateral if $ α+ωβ +ω^2γ = 0 $ or $ α+ω^2β +ωγ = 0 $.
I understand the main idea of rotating the vertices of the triangle by angles that are multiples of $ 60^{\circ}$ and to somehow end up at the position that I started with. However, I still can't figure out how to properly construct that equation for it to be equal to $ 0 $.
Any hints?
HINT 1
Knowing that multiplying a vector by $\omega$ rotates that vector anticlockwise by $\frac{2\pi}{3}$, the triangle is equilateral with $\alpha, \beta, \gamma$ in anticlockwise order if and only if $$\gamma-\beta=(\beta-\alpha)\omega$$
Whereas the triangle is equilateral with $\alpha,\beta,\gamma$ in clockwise order if and only if $$(\gamma-\beta)\omega=\beta-\alpha$$
HINT 2
$\omega^3=1$ and $1+\omega+\omega^2=0$