Question

Consider a quadratic equation $a{z}^2+bz+c=0$, where a, b, c are complex numbers, then the condition that equation has one purely imaginary root is,

• A. $(c\overline{a}-a\overline{c}{)}^2=-(b\overline{c}+c\overline{b}\left)\right(a\overline{b}+\overline{a}b)$
• B. $(c\overline{a}+a\overline{c}{)}^2=(b\overline{c}+c\overline{b}\left)\right(a\overline{b}+\overline{a}b)$
• C. $(c\overline{a}-a\overline{c}{)}^2=(b\overline{c}-c\overline{b}\left)\right(a\overline{b}-\overline{a}b)$
• D. None of these

Answer 1

The correct option is A $(c\overline{a}-a\overline{c}{)}^2=-(b\overline{c}+c\overline{b}\left)\right(a\overline{b}+\overline{a}b)$

Let $z_1$ (purely imaginary) be a root of the given equation. Then,

$z_1=-{\overline{z}}_1$

$a{z}_1^2+b{z}_1+c=0$ (1)

By taking conjugate of both sides

$\Rightarrow a{z}_1^2-b{z}_1+c=0$

$\Rightarrow \overline{a}{\overline{z}}_1^2+\overline{b}{\overline{z}}_1+\overline{c}=0$

$\Rightarrow \overline{a}{z}_1^2-\overline{b}{z}_1+\overline{c}=0$ (as ${\overline{z}}_1=-z_1)$ (2)

Now Eqs. (1) and (2) must have one common root.

$\therefore (c\overline{a}-a\overline{c}{)}^2=(b\overline{c}+c\overline{b}\left)\right(-a\overline{b}-\overline{a}b)$

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