Consider a quadratic equation az2+bz+c=0 , where a,b,c are complex numbers. The condition that the equation has one purely imaginary root is

Let α (purely imaginary) be a root of the given equation thenα=−α¯ . Alsoaα2+bα+c=0       …. (1)From (1), aα2+bα+c¯ =0¯⇒a¯α¯2+b¯α¯+c¯=0⇒a¯α2−b¯α+c¯=0     …. (2)  [∵z=−z¯]Solving (1) and (2) simultaneously, we get α2bc¯+cb¯=αca¯−ac¯=1−ab¯−a¯bEliminating α, we get (bc¯+cb¯)(ab¯+a¯b)+(ca¯−ac¯)2=0Note: Using sum of roots and product of roots is of no use here