Question

If roots of the equation $z^2+\alpha z+\beta =0(\alpha ,\beta \in C)$ are real, then value of $\left(Im\right(\beta ){)}^2+(Im\left(\alpha \right)\left)\right(Im\left(\alpha \overline{\beta }\right))$ is

Note: $Im\left(z\right)$ denotes imaginary part of $z$, where $z$ is a complex number.

• A. $0$
• B. $1$
• C. $-1$
• D. $i$

Answer 1

The correct option is A $0$

$z^2+\alpha z+\beta =0$ ....(1)
where, $z^2+\alpha z+\beta =0$
Since, $z$ is purely real.

$\therefore z=\overline{z}$
Taking conjugate to equation (1), we get

${\overline{z}}^2+\overline{\alpha }\overline{z}+\overline{\beta }=0$

$\because \overline{z_1+z_2}=\overline{z_1}+\overline{z_2}$
$\Rightarrow z^2+\overline{\alpha }z+\overline{\beta }=0$ ......[$z=\overline{z}$ ] ....(2)
Solving (1) and (2) for $z^2$ and $z$, we get
$z^2=\frac{-(\alpha \overline{\beta }-\overline{\alpha }\beta )}{\alpha -\overline{\alpha }}=\frac{-Im\left(\alpha \overline{\beta }\right)}{Im\left(\alpha \right)}$ ....(3)
$z=\frac{-(\beta -\overline{\beta })}{\alpha -\overline{\alpha }}=\frac{-Im\left(\beta \right)}{Im\left(\alpha \right)}$ ....(4)

From (3) and (4), we get
$(Im\beta {)}^2+(Im\alpha \left)\right(Im\left(\alpha \overline{\beta }\right))=0$
Ans: A