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I need some help proving this, I've seen it proven in the other direction (prove the formula if it is an equilateral) but cant figure out how to prove it this way around.

Given three complex numbers $z_1, z_2, z_3$ prove that the points $z_1, z_2, z_3$ are vertices of an equilateral triangle in $\Bbb C$, if $$z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_1z_3 + z_2z_3$$

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Three points $z_1,z_2,z_3$ are vertices of an equilateral triangle $$\iff \frac{z_3-z_1}{z_2-z_1}=\cos(\pm 60^\circ)+i\sin(\pm 60^\circ)=\frac{1\pm\sqrt 3i}{2}$$ $$\iff 2z_3-z_1-z_2=\pm\sqrt 3i(z_2-z_1)$$ $$\iff (2z_3-z_1-z_2)^2=(\pm\sqrt 3i(z_2-z_1))^2$$ $$\iff z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1.$$

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