Condition for 3 complex numbers to represent an equilateral triangle $z_1$, $z_2$, 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.
 A: 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.
A: 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 $\frac{1+2\omega^2}3$
