Algebraic equations and polynomials

Problem:

Consider the quadratic equation

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

 

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