Question

If $\frac{l{z}_1}{m{z}_2}$ is purely imaginary number, then $\left|\frac{\lambda z_1+\mu z_2}{\lambda z_1-\mu z_2}\right|$ is equal to

(Given: $l$, $m$ , $\lambda$, $\mu$ are real numbers)

• A. $\frac{l}{m}$
• B. $\frac{\lambda }{\mu }$
• C. $\frac{-\lambda }{\mu }$
• D. $1$

Answer 1

The correct option is D $1$

Let $\frac{l{z}_1}{m{z}_2}=0+iy\Rightarrow \frac{z_1}{z_2}=\frac{m}{l}iy$
$=|\frac{\lambda z_1+\mu z_2}{\lambda z_1-\mu z_2}|$ dividing numerator & denominator by $z_2$

$=\mid \frac{\frac{\lambda z_1}{z_2}+\mu }{\frac{\lambda z_1}{z_2}-\mu }|$

$=|\frac{\frac{\lambda m}{l}iy+\mu }{\frac{\lambda miy}{l}-\mu }|$

$=\frac{\sqrt{{\mu }^2+(\frac{\lambda m}{l}y{)}^2}}{{\sqrt{(-\mu {)}^2+\frac{\lambda m}{l}y}}^2}$
$=1$