Question

A complex number z is said to be unimodular, if $|z|eq1$. If $z_1$ and $z_2$ are complex numbers such that $\frac{z_1-{2z}_2}{2-z_1{\stackrel{-}{z}}_2}$ 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 $2$
• D. circle of radius $\sqrt{2}$

Answer 1

The correct option is C circle of radius $2$

We have,

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

$\Rightarrow |z_1-2z_2|=|2-z_1\overline{z_2}|$

On squaring both side and we get,

$(z_1-2z_2)(\overline{z_1}-2\overline{z_2})=(2-z_1\overline{z_2})(2-\overline{z_1}{z}_2)$

$\Rightarrow z_1\overline{z_1}-2z_2\overline{z_1}-2z_1\overline{z_2}+4\overline{z_2}{z}_2=4-2z_2\overline{z_1}-2z_1\overline{z_2}+\overline{z_1}{z}_2{z}_1\overline{z_2}$

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

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

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

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

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

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

Now,

$\left|z_2\right|\ne 0$

Then

$\left|z_1\right|=2$

Hence, the radius of circle is $2$.

This is the answer.