Question

Let $z_1$ and $z_2$ be two imaginary roots of $z^2+pz+q=0$, where $p$ and $q$ are real. The points $z_1,z_2$ and origin form an equilateral triangle if

• A. $p^2>3q$
• B. $p^2<3q$
• C. $p^2=3q$
• D. $p^2=q$

Answer 1

The correct option is C $p^2=3q$

Since $p,q\in R$,
$\therefore z_1,z_2$ are in conjugate pairs.

Let $z_1=\alpha +i\beta$ and $z_2=\alpha -i\beta$
$z_1+z_2=2\alpha =-p$
$z_1{z}_2={\alpha }^2+{\beta }^2=q$

Solving the two equations, we get
$z_1\equiv A(\frac{-p}{2},\sqrt{\frac{4q-p^2}{4}})$
$z_2\equiv B(\frac{-p}{2},-\sqrt{\frac{4q-p^2}{4}})$
$O\equiv (0,0)$

$z_1,z_2$ and origin form an equilateral triangle.
So, $OA=AB$
$\Rightarrow \frac{p^2}{4}+\frac{4q-p^2}{4}={(\sqrt{\frac{4q-p^2}{4}}+\sqrt{\frac{4q-p^2}{4}})}^2$
$\Rightarrow q=4q-p^2$
$\Rightarrow p^2=3q$