Question

A complex number $z$ is said to be unimodular if $|z|=1$ . Let, $z_1$ and $z_2$ are complex numbers such that $\frac{{\text{z}}_1-2z{}_2}{2-{\text{z}}_1{\bar{\text{z}}}_2}$ is unimodular and $z_2$ is not unimodular, then the point $z_1$ lies on a

• A. circle of radius $\sqrt{2}$
• B. straight line parallel to $x$ -axis
• C. straight line parallel to $y$ -axis
• D. circle of radius $2$

Answer 1

Answer: (D)

Given, $\frac{{\text{z}}_1-2z{}_2}{2-{\text{z}}_1{\bar{\text{z}}}_2}$ is unimodular
$\Rightarrow \left|\frac{{\text{z}}_1-2{\text{z}}_2}{2-{\text{z}}_1{\bar{\text{z}}}_2}\right|=1$
$\Rightarrow \left|{\text{z}}_1-2{\text{z}}_2\right|=\left|2-{\text{z}}_1{\bar{\text{z}}}_2\right|$
Squaring both the sides, we get,
${\left|{\text{z}}_1-2z{}_2\right|}^2={\left|2-{\text{z}}_1{\bar{\text{z}}}_2\right|}^2$
$\Rightarrow \left({\left(\text{z}\right)}_1-2{\left(\text{z}\right)}_2\right)\left({\left(\bar{\text{z}}\right)}_1-2{\left(\bar{\text{z}}\right)}_2\right)=\left(2-{\left(\text{z}\right)}_1{\left(\bar{\text{z}}\right)}_2\right)\left(2-{\left(\bar{\text{z}}\right)}_1{\left(\text{z}\right)}_2\right)$
$\left(\because {\left(\left|\text{z}\right|\right)}^2=\text{z }\bar{\text{z}}\right)$
$\Rightarrow {\text{z}}_1{\bar{\text{z}}}_1-2{\text{z}}_1{\bar{\text{z}}}_2-2{\bar{\text{z}}}_1{\text{z}}_2+4{\text{z}}_2{\bar{\text{z}}}_2$
$=4-2{\bar{\text{z}}}_1{\text{z}}_2-2{\text{z}}_1{\bar{\text{z}}}_2+{\text{z}}_1{\bar{\text{z}}}_1{\text{z}}_2{\bar{\text{z}}}_2$
$\Rightarrow {\left|{\text{z}}_1\right|}^2+4{\left|{\text{z}}_2\right|}^2=4+{\left|{\text{z}}_1\right|}^2{\left|{\text{z}}_2\right|}^2$
$\Rightarrow {\left|{\text{z}}_1\right|}^2-4+4{\left|{\text{z}}_2\right|}^2-{\left|{\text{z}}_1\right|}^2{\left|{\text{z}}_2\right|}^2=0$
$\Rightarrow \left({\left(\left|{\left(\text{z}\right)}_1\right|\right)}^2-4\right)\left(1-{\left(\left|{\left(\text{z}\right)}_2\right|\right)}^2\right)=0$
$\Rightarrow \left|{\text{z}}_1\right|=2\text{or }\left|{\text{z}}_2\right|=1$
Given, $z_2$ is not unimodular
$\therefore \left|{\text{z}}_1\right|=2$
$\therefore$ Point $z_1$ lies on a circle of radius $2$ .