Consider a quadratic equation $az^2+bz+c=0$ where a,b,c are complex numbers. Prove that the equation has one purely imaginary root is given ... Problem : 
Consider a quadratic equation $az^2+bz+c=0$ where a,b,c are complex numbers. Prove that the condition such that the equation has one purely imaginary root is given by $(b\overline{c}+c\overline{b})(a\overline{b}+\overline{a}b)+(c\overline{a}-a\overline{c})^2=0$
I am not getting any clue on this how to proceed such equation in complex numbers. Please suggest thanks a lot.
 A: Let $r$ be the purely imaginary root, so that $\overline{r} = -r$. We have these conjugate equations:
$$a r^2 + b r^1 + c = 0 \qquad \text{and} \qquad \overline{a} r^2 - \overline{b} r^1 + \overline{c} = 0$$
Here's a neat trick: Move the exponents on $r$ to subscripts ...
$$a r_2 + b r_1 + c = 0 \qquad \text{and} \qquad \overline{a}r_2 - \overline{b} r_1 + \overline{c} = 0$$
... and solve the linear system for them ...
$$
r_1 =  \frac{a\overline{c}-\overline{a}c}{a\overline{b}+\overline{a}b} \qquad\qquad
r_2 = -\frac{b\overline{c}+\overline{b}c}{a\overline{b}+\overline{a}b}
$$
(Dealing with the case of $a\overline{b}+\overline{a}b=0$ is left as an exercise to the reader.) Now, return the subscripts to their positions as exponents ...
$$
r^1 =  \frac{a\overline{c}-\overline{a}c}{a\overline{b}+\overline{a}b} \qquad\qquad
r^2 = -\frac{b\overline{c}+\overline{b}c}{a\overline{b}+\overline{a}b}
$$
... and simply observe the deep algebraic truth, $r^2 = (r^1)^2$ ...
$$\left(\;\frac{a\overline{c} -\overline{a}c}{a\overline{b}+\overline{a}b}\;\right)^2 = -\frac{b\overline{c}+\overline{b}c}{a\overline{b}+\overline{a}b}$$
Squaring and clearing fractions, we have
$$(a\overline{c} -\overline{a}c)^2 = -(a\overline{b}+\overline{a}b)( b\overline{c}+\overline{b}c)$$
