Question

Consider a quadratic equation $a{z}^2+bz+c=0$, where $a,b,c$ are complex numbers and if equation has two purely imaginary roots, then which of the given is true?

• A. $a\overline{b}$ is purely imaginary
• B. $b\overline{c}$ is purely imaginary
• C. $c\overline{a}$ is purely real
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A $a\overline{b}$ is purely imaginary
C $c\overline{a}$ is purely real

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

Let, $z_1$ and $z_2$ be the roots of the above equation.

Since both the roots are purely imaginary.

$\therefore z_1=-\overline{z_1}\&z_2=-\overline{z_2}$ ...$\left(1\right)$

Sum of the roots of quadratic equation is $z_1+z_2=\frac{-b}{a}$ ...$\left(2\right)$

Product of the roots is $z_1{\times z}_2=\frac{c}{a}$ ...$\left(3\right)$

Taking conjugate of eq. $\left(2\right)$ and eq. $\left(3\right)$ we get,

$\overline{z_1}+\overline{z_2}=\frac{-\overline{b}}{\overline{a}}$ ...$\left(4\right)$

$\overline{z_1{\times z}_2}=\frac{\overline{c}}{\overline{a}}$ ...$\left(5\right)$

Using eq. $\left(1\right)$, eq. $\left(4\right)$ and eq. $\left(5\right)$ we get,

$a\overline{b}=-\overline{a}b$

$⟹a\overline{b}=-\overline{a\overline{b}}$

$\therefore a\overline{b}$ is purely imaginery.

$c\overline{a}=\overline{c}a$

$⟹c\overline{a}=\overline{c\overline{a}}$

$\therefore c\overline{a}$ is purely real.

Ans: $A$ and $C$