Question

A complex number $z$ is said to be unimodular if $|z|=1$. Let $z_1$ and $z_2$ be two complex numbers such that $\frac{z_1-2z_2}{2-z_1\overline{z_2}}$ is unimodular and $z_2$ is not unimodular. Then the point $z_1$ lies on a

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

Answer 1

The correct option is D Circle of radius $2$

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