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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.

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  • $\begingroup$ This might help. $\endgroup$ – Emin Feb 12 at 12:54
  • $\begingroup$ Isn't there a non-revealed assumption? Why am I asking? Consider $z_1=z_2.$ From the equation we get $z_3=z_1$ which is a degenerate equilateral triangle, or $z_3=-z_1$ which doesn't define a triangle at all. $\endgroup$ – user376343 Feb 13 at 22:50
  • $\begingroup$ I'm sorry, they are distinct too. $\endgroup$ – Parallelism Alert Feb 14 at 9:30

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