Q. A real value of $x$ will satisfy the equation $\left( \frac{3 - 4ix}{3 + 4ix} \right) = \alpha - i\beta\ (\alpha,\beta\ are\ real)$, if
A) $\alpha^{2} - \beta^{2} = - 1$
B) $\alpha^{2} - \beta^{2} = 1$
C) $\alpha^{2} + \beta^{2} = 1$
D) $\alpha^{2} - \beta^{2} = 2$
Explanation
(c)
Given, $\alpha - i\beta = \left( \frac{3 + i( - 4x)}{3 + i(4x)} \right)$
$\Rightarrow \ \ \ \ \ \ \left| \alpha + i( - \beta) \right| = \left| \frac{3 + i( - 4x)}{3 + i(4x)} \right|$
$= \frac{|3 + i( - 4x)|}{|3 + i(4x)|}$
$\Rightarrow \ \ \ \ \alpha^{2} + \beta^{2} = \frac{9 + 16x^{2}}{9 + 16x^{2}\ }$
$\Rightarrow \ \ \ \ \ \ \ \ \alpha^{2} + \beta^{2} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.