Consider a quadratic equation `az^(2)bz+c=0,` where a,b and c are complex numbers.
the condition that equation has two purely imaginary roots, is

Options

A. `(bar(a))/a=(bar(b))/b=(bar(c))/c`
B. `(bar(a))/a=-(bar(b))/b=(bar(c))/c`
C. `(bar(a))/a=(bar(b))/b=-(bar(c))/c`
D. `-(bar(a))/a=(bar(b))/b=(bar(c))/c`

Answer

B

Explanation

Let `alpha` be one root of equation then `bar(alpha)=-alpha`
Given equation is `az^(2)+bz+c=0`
`:.aalpha^(2)+balpha+c=0`
Taking conjugate we get `:.bar(a).bar(alpha)^(2)+bar(b).bar(alpha)+bar(c)=0`
`bar(a).alpha^(2)-bar(b).alpha+bar(c)=0` so `alpha.beta` are roots of equation
`az^(2)+bz+c=0` and `bar(a)z^(2)-bar(b)z+bar(c)=0`
So `(bar(a))/a=-(bar(b))/b=(bar(c))/c`
if `|alpha|=1` then `bar(alpha)=1/(alpha).` As `aalpha^(2)+balpha+c=0`..........1
`impliesbar(a)bar(alpha)^(2)+bar(b)bar(alpha)+bar(c)=0`
`implies(bar(a))/(alpha^(2))+(bar(b))/(alpha)+bar(c)=0`
So `bar(c).alpha^(2)+bar(b).alpha+bar(a)=0`.........2
Now applying condition for one common root for 1 and 2 we get `(a bar(b)-b bar(c))(b bar(a)-c bar(b))=(a bar(a)-c bar(c))^(2)`