Let: $z_1 =e^{ia} ; z_2 = e^{ib}; z_3 = e^{ic}$
$ z_1 +z_2 = e^{i\frac{a+b}{2}}*(e^{i\frac{a-b}{2}} + e^{-i\frac{(a-b)}{2}}) = e^{i\frac{a+b}{2}}*2*cos(\frac{a-b}{2}) = -z_3 $
=> $|2*cos(\frac{a-b}{2})| = |-z_3| = |z_3| = 1$ ,
If $ cos(\frac{a-b}{2}) =\frac{1}{2} $ -> $a = b \pm \frac{2\pi}{3}$ $mod(2\pi)$
here without loss of generality you can assume a= b+ $\frac{2\pi}{3}$ $ mod(2\pi)$ (the other case is the same)
you get : $\frac{a+b}{2} = c+\pi$ $ mod(2\pi)$ -> b+ $\frac{\pi}{3} = c + \pi$ $ mod(2\pi)$ -> $ b = c + \frac{2\pi}{3} $ $ mod(2\pi)$
You get your equilateral triangle, since you proved that you can rotate of $\frac{2\pi}{3}$ to pass from one point to another. The other cases are exactly the same.
As for 2) , I would use : $z= e^{ia}$
$z^2 - z = e^{2ia} - e^{ia}$ = $e^{\frac{3}{2}ia}*2i*sin(\frac{a}{2}) $ = $ e^{(\frac{3}{2}a + \frac{\pi}{2})i}*2*sin(\frac{a}{2}) $. The sign of the sin is the only thing you have take into account to evaluate correctly the argument. If it is negative, you add $\pi$, else you already have your argument
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).
Best Answer
Hint: Consider the expression $a = z_1 + \omega z_2 + \omega^2 z_3$. It is invariant under translation ($(z_1,z_2,z_3) \mapsto (z_1+z,z_2+z,z_3+z)$) since $1+\omega+\omega^2=0$, and it is homogeneous, that is $a(zz_1,zz_2,zz_3) = za(z_1,z_2,z_3)$. Since all equilateral triangles can be transformed to one another using translation and rotation/scaling (the operation corresponding to multiplying all points by some complex number), it is enough to prove the result for some fixed equilateral triangle. Choose one and do the math.