Prove that $z_1,z_2,z_3 \in \Bbb C$ , distinct, with equal, non-zero modulus, are vertices of equilateral triangle if and only if $\sum_{cyc}|z_1-z_2|(z_1+z_2) = 0 $
I tried dividing by $z_3|z_3|\ne0$ and I obtained $$|a-b|(a+b)+|a-1|(a+1)+|b-1|(b+1)=0\text{, where }a=\frac{z_1}{z_3}, b=\frac{z_2}{z_3}, |a|=|b|=1$$
The equivalent condition for the triangle to be equilateral is: $$z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_1z_3\text{, equivalent to} $$ $$a^2+b^2+1= ab+a+b $$
The last condition could be written as $(a-b)^2+(a-1)(b-1)=0$
Surely, I need to prove that the equation in the title implies that the triangle is equilateral. The implication from the fact that the triangle is equilateral, resulting the equation, is obvious.