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

Let $\alpha$ (purely imaginary) be a root of the given equation then$\alpha =-\overline{\alpha }$ . Also$a{\alpha }^2+b\alpha +c=0$       …. (1)

From (1), $\overline{a{\alpha }^2+b\alpha +c}$ $=\overline{0}$

$\Rightarrow \overline{a}{\overline{\alpha }}^2+\overline{b}\overline{\alpha }+\overline{c}=0\Rightarrow \overline{a}{\alpha }^2-\overline{b}\alpha +\overline{c}=0$     …. (2)  $\left[\because z=-\overline{z}\right]$

Solving (1) and (2) simultaneously, we get $\frac{{\alpha }^2}{b\overline{c}+c\overline{b}}=\frac{\alpha }{c\overline{a}-a\overline{c}}=\frac{1}{-a\overline{b}-\overline{a}b}$

Eliminating $\alpha ,$ we get $\left(b\overline{c}+c\overline{b}\right)\left(a\overline{b}+\overline{a}b\right)+{\left(c\overline{a}-a\overline{c}\right)}^2=0$

Note: Using sum of roots and product of roots is of no use here