Question

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

Answer 1

Let $P=\frac{2z_1}{3z_2}$

For P being purely imaginary

$P+\overline{P}=0$

$\frac{z_1}{z_2}+\frac{\overline{z_1}}{\overline{z_2}}=0$

$\frac{z_1\overline{z_2}+z_2\overline{z_1}}{z_2\overline{z_2}}=0$

$z_1\overline{z_2}+z_2\overline{z_1}=0$

Now,

${\left|\frac{z_1-z_2}{z_1+z_2}\right|}^2=\left[\frac{z_1-z_2}{z_1+z_2}\right]\left(\frac{\overline{z_1}-\overline{z_2}}{\overline{z_1}+\overline{z_2}}\right)$

${\left|\frac{z_1-z_2}{z_1+z_2}\right|}^2=\frac{z_1\overline{z_1}-z_2\overline{z_1}-\overline{z_2}{z}_1+z_2\overline{z_2}}{z_1\overline{z_1}+\overline{z_1}{z}_2+z_1\overline{z_2}+z_2\overline{z_2}}$

$=\frac{{\left|z_1\right|}^2+{\left|z_2\right|}^2}{{\left|z_1\right|}^2+{\left|z_2\right|}^2}=1$

Therefore, $|\frac{z_1-z_2}{z_1+z_2}|=1$