, 2011

Let $a$, $b$, $c$ be complex numbers such that $a + b + c = 0$ and $ab + bc + ca = 0$.

Let us consider the elementary symmetric polynomials:

- $s_1 = a + b + c = 0$
- $s_2 = ab + bc + ca = 0$
- $s_3 = abc$

The roots $a$, $b$, $c$ are the roots of the cubic equation:
$$x^3 - s_1 x^2 + s_2 x - s_3 = 0$$
With $s_1 = 0$ and $s_2 = 0$, this becomes:
$$x^3 - s_3 = 0$$
So $a$, $b$, $c$ are the three cube roots of $s_3$.

But $a + b + c = 0$ implies that the sum of the three roots is zero, which is only possible if $a$, $b$, $c$ are equally spaced on the complex plane around the origin, i.e., they form the vertices of an equilateral triangle centered at the origin.

Let $a = r e^{i heta}$, $b = r e^{i( heta + 2\pi/3)}$, $c = r e^{i( heta + 4\pi/3)}$ for some $r > 0$ and $heta \in \mathbb{R}$.

Then $|a| = |b| = |c| = r$.

Therefore, $|a| = |b| = |c|$.