showing certain vertices form an equilateral triangle Suppose $a,b,c$ lie in the unit circle of the complex plane and satisfy $a+b+c = 0$. Then, $a,b,c$ form the vertices of an equilateral triangle.
Try
So, I want to show that $|b-a|=|c-a|=|b-c|$. We are given $|a| = |b| = |c| = 1$ and $a+b+c=0$. Since $|z| = z \overline{z} $, we have
$$ (b-a)( \overline{b}-\overline{a} ) = (c-a)( \overline{c} - \overline{a} )$$
which simplifies to
$$ |b| - b \overline{a} - a \overline{b} - |a| = |c| - c \overline{a} - a \overline{c} - |a| $$
Hence,
$$ c \overline{a} + a \overline{c} - b \overline{a} - a \overline{b} = 0$$
$$ a( \overline{c} - \overline{b} ) + \overline{a} ( c-b) = 0 $$
Here I am stuck. Perhaps this is the wrong approach?

Without loss of generality you can assume that $a=1$, because the centre of mass of the three points, the origin, is invariant under their rotation about it. This implies that $c = \overline b$, and that the real component of $b$ and $c$ is $-1/2$.
