Question

If $z_1$ and $z_2$ are complex numbers such that $\frac{2z_1}{3z_2}$ is purely imaginary number, then find $\left|\frac{z_1-z_2}{z_1+z_2}\right|$

Answer 1

Given, $z_1$ and $z_2$ are complex numbers, such that $\frac{2z_1}{3z_2}$ is purely imaginary.

$\therefore$ $\frac{2z_1}{3z_2}=yi,y\in R\Rightarrow \frac{z_1}{z_2}=\frac{3}{2}yi\left[2\right]$

Now $\left|\frac{z_1-z_2}{z_1+z_2}\right|=\left|\frac{\frac{z_1}{z_2}-1}{\frac{z_1}{z_2}+1}\right|=\left|\frac{\frac{3}{2}yi-1}{\frac{3}{2}yi+1}\right|=\left|\frac{3yi-2}{3yi+2}\right|=\frac{\sqrt{9y^2+4}}{\sqrt{9y^2+4}}=1\left[2\right]$