Question

If $\frac{2z_1}{3z_2}$ is a purely imaginary, then $\left|\frac{z_1-z_2}{z_1+z_2}\right|$ is:

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{4}{9}$
• D. $1$
• E. $\frac{9}{4}$

Answer 1

Answer: (D)

$\because$ $\frac{2z_1}{3z_2}$ is purely imaginary
$\Rightarrow$ $\frac{2z_1}{3z_2}=ki$ where $k\in R$
$\Rightarrow$ $\frac{z_1}{z_2}=\frac{3}{2}ki$
Now, $\left|\frac{z_1-z_2}{z_1+z_2}\right|=\left|\frac{\frac{z_1}{z_2}-1}{\frac{z_1}{z_2}+1}\right|=\left|\frac{3ki-2}{3ki+2}\right|$
$=1$