If $z_1'=z_1z_2-z_3^2,~z_2'=z_2z_3-z_1^2,~z_3'=z_3z_1-z_2^2$ are the affixes of the vertices of an equilateral triangle, then $z_1,~z_2,~z_3$ are also the affixes of an equilateral triangle and vice-versa.
My try:
Denote $A(z_1),~B(z_2),~C(z_3)$ and $A'(z_1'),~B'(z_2'),C'(z_3')$
$\triangle ABC$ – equilateral $\Leftrightarrow$ $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_1z_3$ (1)
$\triangle A'B'C'$ – equilateral $\Leftrightarrow$ $z_1'^2+z_2'^2+z_3'^2=z_1'z_2'+z_2'z_3'+z_1'z_3'$ (2)
Working on (2) it follows that
$z_1^4+z_2^4+z_3^4-3z_1z_2z_3(z_1+z_2+z_3)+z_1^3z_2+z_1^3z_3+z_2^3z_1+z_2^3z_3+z_3^3z_1+z_3^3z_2=0$
Which is factorable:
$(z_1+z_2+z_3)^2(z_1^2+z_2^2+z_3^3-z_1z_2-z_2z_3-z_3z_1)=0$
Case I — $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1\Rightarrow$ (1)
With this we have proved implication (1) $\Rightarrow$ (2) and also proved implication (2) $\Rightarrow$ (1) for this case.
Case II — $z_1+z_2+z_3=0$
For this case, implication (2) $\Rightarrow$ (1) is not immediately proved as in case I.
Geometric meaning of the equation:
Let G be the center of gravity and O be the center of the circle circumscribed to $\triangle ABC$. Then, G coincides with the origin. If we can somehow prove that O also coincides with the origin, that is, $|z_1|=|z_2|=|z_3|$, then it means that $\triangle ABC$ – equilateral.
But how can I prove it?
Best Answer
If $z_1 + z_2 + z_3 = 0$, then $\Delta A'B'C'$ is degenerate (it reduces to a single point). To see this, note that
$$z_1' - z_2' = z_1 z_2 - z_2z_3 + z_1^2 - z_3^2 = z_2(z_1-z_3) + (z_1-z_3)(z_1+z_3) $$
which is just
$$ (z_1 - z_3)(z_1+z_2+z_3) = 0.$$
Thus we get $z_1' = z_2' = z_3'$, so i guess that you can exclude this case to make your proof work (I don't think a point would be considered an equilateral triangle).