Prove that the distinct complex numbers \(z_1, z_2, z_3\) are the vertices of an equilateral triangle if and only if \[ z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1} .\]
Solution
Translate so that \(z_1=0\), and put \[ a=z_2-z_1,\qquad b=z_3-z_1. \] The displayed equation is translation-invariant and becomes \[ a^2+b^2=ab. \] Since the three points are distinct, \(a\ne0\). Dividing by \(a^2\) and writing \(t=b/a\) gives \[ t^2-t+1=0, \] so \[ t=e^{i\pi/3}\quad\text{or}\quad t=e^{-i\pi/3}. \] Thus \(|b|=|a|\) and the angle between \(a\) and \(b\) is \(\pi/3\). Consequently \[ |a-b|=|a|=|b|, \] so the three vertices form an equilateral triangle.
Conversely, if the triangle is equilateral, after the same translation and division by \(a\ne0\) we have \(b/a=e^{\pm i\pi/3}\), which satisfies \(t^2-t+1=0\). Reversing the algebra gives the displayed identity.