Question

If $\left|\frac{(z_1-2z_2)}{(2-z_1{\overline{z}}_2)}\right|=1$ and $|z_2|\ne 1,$ where $z_1$ and $z_2$ are complex numbers, then the value of $|z_1|$ is

• A. 1
• B. $-1$
• C. 2
• D. $-2$

Answer 1

Answer: (C)

Given, $\left|\frac{(z_1-2z_2)}{(2-z_1{\overline{z}}_2)}\right|=1,|z_2|\ne 1$
$\Rightarrow$ $|z_1-2z_2{|}^2=|2-z_1\overline{z}{|}^2$
$\Rightarrow$ $(z_1-2z_2)(\overline{z_1-2z_2})=(2-z_1{\overline{z}}_2)(\overline{2-z_1{\overline{z}}_2})$
$\Rightarrow$ $(z_1-2z_2)({\overline{z}}_1-2{\overline{z}}_2)=(2-z_1{\overline{z}}_2)(2-z_1{\overline{z}}_2)$
$\Rightarrow$ $z_1{\overline{z}}_1-2z_1{\overline{z}}_2-2{\overline{z}}_1{z}_2+4z_2{\overline{z}}_2$ $=4-2{\overline{z}}_1{z}_2-2z_1{\overline{z}}_2+z_1{\overline{z}}_1{z}_2{\overline{z}}_2$
$\Rightarrow$ $|z_1{|}^2+4|z_2{|}^2=4+|z_1{|}^2|z_2{|}^2$
$\Rightarrow$ $|z_1{|}^2+4|z_2{|}^2-4-|z_1{|}^2|z_2{|}^2=0$
$\Rightarrow$ $|z_1{|}^2(1-|z_2{|}^2)-4(1-|z_2{|}^2)=0$
$\Rightarrow$ $(1-|z_2{|}^2)(|z_2{|}^2-4)=0$
$\Rightarrow$ $1-|z_2{|}^2=0|z_1{|}^2-4=0$
$\Rightarrow$ $|z_2|\ne 1,|z_1|=2$