Question

Consider a quadratic equation $a{z}^2+bz+c=0$, where a, b, c are complex numbers. If the equation has two purely imaginary roots, then which of the following is not 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

The correct option is D None of these

Let $z_1$ and $z_2$ be two purely imaginary roots. Then,

${\overline{z}}_1=-z_1,{\overline{z}}_2=-z_2$

Now, $\underset{–}{a{z}^2+bz+c}=0$ (1)

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

Taking conjugate of both sides

or $\overline{a}{\overline{z}}^2+\overline{b}\overline{z}+\overline{c}=0$

or $\overline{a}{z}^2-\overline{b}z+\overline{c}=0$ (2)

Equations (1) and (2) must be identical as their roots are same.

$\therefore \frac{a}{\overline{a}}=-\frac{b}{\overline{b}}=\frac{c}{\overline{c}}$

$\Rightarrow a\overline{c}=\overline{a}c,a\overline{b}+\overline{a}b=0$ and $b\overline{c}+\overline{b}c=0$

Hence, $a\overline{c}$ is purely real and $a\overline{b}$ and $b\overline{c}$ are purely imaginary.

Hence option ${}^{\prime }{D}^{\prime }$ is the answer.