Question

A complex number z is said to be unimodular, if $|z|=1$. If and $z_1$ and $z_2$ are complex numbers such that $\frac{z_1-2z_2}{2-\left(z_1\overline{z_2}\right)}$ 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 X-axis
• C. Circle of radius 2
• D. Circle of radius $\sqrt{2}$

Answer 1

The correct option is C Circle of radius 2

If Z is unimodular, then |z|=1. Also, use property of modulus i.e $z\overline{z}=|z{|}^2$
Given, $z_2$ is not unimodular i.e. $|z_2|\ne 1$ and $\frac{z_1-2z_2}{2-z_1\overline{z_2}}$ is unimodular
$\Rightarrow \left|\frac{z_1-2z_2}{2-z_1\overline{z_2}}\right|=1\Rightarrow |z_1-2z_2{|}^2=|2-z_1\overline{z_2}{|}^2\Rightarrow (z_1-2z_2)(\overline{z_1}-2\overline{z_2})=(2-z_1\overline{z_2})(2-{\overline{z}}_1{z}_2)\Rightarrow |z_1{|}^2+4|z_2{|}^2-2\overline{z_1}{z}_2-2z_1\overline{z_2}=4+|z_1{|}^2|z_2{|}^2-2\overline{z_1}{z}_2-2z_1\overline{z_2}\Rightarrow (|z_2{|}^2-1)(|z_1{|}^2-4)=0\because |z_2|\ne 1\therefore |z_1|=2$
Let $z_1=x+iy\Rightarrow x^2+y^2=(2{)}^2$
$\therefore$ Point $z_1$ lies on a circle of radius 2.