Condition for 3 complex numbers to represent an equilateral triangle

Question

$z_1$, $z_2$, and $z_3$ are 3 complex numbers. Prove that if they represent the vertices of an equilateral triangle then $z_1 + \omega z_2 + \omega^2 z_3 = 0$ where $\omega$ is a 3rd root of unity. Any help would be thoroughly appreciated.

Answer 1

Hint: Consider the expression $a = z_1 + \omega z_2 + \omega^2 z_3$. It is invariant under translation ($(z_1,z_2,z_3) \mapsto (z_1+z,z_2+z,z_3+z)$) since $1+\omega+\omega^2=0$, and it is homogeneous, that is $a(zz_1,zz_2,zz_3) = za(z_1,z_2,z_3)$. Since all equilateral triangles can be transformed to one another using translation and rotation/scaling (the operation corresponding to multiplying all points by some complex number), it is enough to prove the result for some fixed equilateral triangle. Choose one and do the math.

Answer 2

Yuval's answer cannot be improved on for elegance, but a slightly messier solution may be worth a footnote

(taking note of @ccorn's shrewd insight we'll here explicitly number the vertices anticlockwise, i.e. co-$\omega$-wise). it is easy to see geometrically that the equilateral symmetry can be expressed in terms of invariance of the figure under the action of $C_3$ represented as rotations about the centroid of the triangle.

since multiplication by $\omega$ effects an anti-clockwise rotation through $\frac{2\pi}3$ for the triangle to be equilateral requires: $$ \left(z_1 - \frac{z_1+z_2+z_3}3 \right)\omega = z_2 - \frac{z_1+z_2+z_3}3 $$ i.e. $$ (2\omega+1)z_1-(2+\omega)z_2+(1-\omega)z_3 = 0 $$ and the result follows on multiplication by $\frac{1+2\omega^2}3$