# Algebraic equations and polynomials

Problem:

$a^{2}z^{2}+abz+c^{2}=0$

where a, b, $c \in C^{*}$ and denote by $z_{1}$, $z_{2}$ its roots. Prove that if $\frac{b}{c}$ is a real number then $|z_{1}|=|z_{2}|$ or $\frac{z_{1}}{z_{2}} \in \Re$.

Proof:

Let $t = \frac{b}{c} \in \Re$. Then, $b=tc$ and

$\delta = (ab)^{2}-4a^{2}.c^{2}=a^{2}c^{2}(t^{2}-4)$.

If $|t| \geq 2$, the roots of the equation are

$z_{1,2}=-\frac{-tac \pm ac \sqrt{t^{2}-4}}{2a^{2}}=\frac{c}{2a}(-t \pm \sqrt{t^{2}-4})$ and it is obvious that $\frac{z_{1}}{z_{2}}$ is a real  number.

If $|t|<2$,, the roots of the equation are

$z_{1,2}=\frac{c}{2a}(-t \pm i \sqrt{4-t^{2}})$

hence, $|z_{1}|=|z_{2}|=\frac{|c|}{|a|}$, as claimed.

QED.

More later,

Nalin Pithwa