Question

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 –

• A. $(b\bar{c} + c\bar{b})(a\bar{b} + \bar{a}b) + (c\bar{a} - a\bar{c})^{2} = 0$
• B. $(b\bar{c} + c\bar{b})(c\bar{a} + a\bar{b}) + (a\bar{b} - \bar{a}b)^{2} = 0$
• C. $(a\bar{b} + \bar{a}b)(c\bar{a} + \bar{c}a) + (b\bar{c} - \bar{b}c)^{2} = 0$
• D. None of these

Answer 1

Answer: (A)

$(b\bar{c} + c\bar{b})(a\bar{b} + \bar{a}b) + (c\bar{a} - a\bar{c})^{2} = 0$

Answer 2

Sol. Let a (purely imaginary) be a root of the given equation then a = –$\bar{\alpha}$.

Also a a² + ba + c = 0 … (1)

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

Ž $\bar{a}{\bar{\alpha}}^{2} + \bar{b}\bar{\alpha} + \bar{c} = 0$Ž$\bar{a}\alpha^{2}–\bar{b}\alpha + \bar{c} = 0$… (2)

[Q z = –$\bar{z}$]

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

Eliminating a, we get$(b\bar{c} + c\bar{b})(a\bar{b} + \bar{a}b) + (c\bar{a} - a\bar{c})^{2} = 0$