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

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

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