JEE Main 2015 — Complex Number Question with Solution

Given, $\frac{z_1-2z_2}{2-z_1{\overline{z}}_2}$  is unimodular. 

$\Rightarrow \left|\frac{z_1-2z_2}{2-z_1{\overline{z}}_2}\right|=1$

$\Rightarrow \left|z_1-2z_2\right|=\left|2-z_1{\overline{z}}_2\right|$

Squaring both the sides, we get,

${\left|z_1-{\mathrm{2z}}_2\right|}^2={\left|2-z_1{z}_2\right|}^2$

$\Rightarrow \left(z_1-2z_2\right)\left({\overline{z}}_1-2{\overline{z}}_2\right)=\left(2-z_1{\overline{z}}_2\right)\left(2-{\overline{z}}_1{z}_2\right)$

$\left(\because {\left|z\right|}^2=z\overline{z}\right)$

$\Rightarrow z_1{\overline{z}}_1-2z_1{\overline{z}}_2-2{\overline{z}}_1{z}_2+4{\overline{z}}_2{\overline{z}}_2$

$=4-2{\overline{z}}_{\mathrm{1 }}{z}_2-2z_{\mathrm{1 }}{\overline{z}}_2+z_{\mathrm{1 }}{\overline{z}}_{\mathrm{1 }}{z}_{\mathrm{2 }}{\overline{z}}_2$

$\Rightarrow {\left|z_1\right|}^2+4{\left|z_2\right|}^2=4+{\left|z_1\right|}^{\mathrm{2 }}{\left|z_2\right|}^2$

$\Rightarrow {\left|z_1\right|}^2-4+4{\left|z_2\right|}^2-{\left|z_1\right|}^{\mathrm{2 }}{\left|z_2\right|}^2=0$

$\Rightarrow \left({\left|z_1\right|}^2-4\right)\left(1-{\left|z_2\right|}^2\right)=0$

$\Rightarrow \left|z_1\right|=2$ or $\left|z_2\right|=1$

 Given, $z_2$ is not unimodular

$\therefore \left|z_1\right|=2$

$\therefore$  Point $z_1$ lies on a circle of radius $2$.