Question

Let $\left|\frac{\overline{z_1}-2\overline{z_2}}{2-z_1\overline{z_2}}\right|=1$ and $|z_2|\ne 1,$ where $z_1$ and $z_2$ are complex numbers. Then $|z_1|$ equals

Answer 1

$\left|\frac{\overline{z_1}-2\overline{z_2}}{2-z_1\overline{z_2}}\right|=1$
$\Rightarrow |\overline{z_1}-2\overline{z_2}{|}^2=|2-z_1\overline{z_2}{|}^2$
$[\because |z_1-z_2{|}^2=|z_1{|}^2+|z_2{|}^2-2Re(z_1\overline{z_2}\left)\right]$
$\Rightarrow |z_1{|}^2+4|z_2{|}^2-2Re\left(\overline{z_1}\overline{\left(\overline{2z_2}\right)}\right)=4+|z_1\overline{z_2}{|}^2-2Re\left(2\right(\overline{z_1\overline{z_2}}\left)\right)$
$[\because \overline{\overline{z}}=z,\overline{z_1{z}_2}=\overline{z_1}\overline{z_2}]$
$\Rightarrow |z_1{|}^2+4|z_2{|}^2-2Re\left(2\overline{z_1}{z}_2\right)=4+|z_1{|}^2|\overline{z_2}{|}^2-2Re\left(2\overline{z_1}{z}_2\right)$
$[\because |\overline{z}|=|z|]$
$\Rightarrow |z_1{|}^2+4|z_2{|}^2=4+|z_1{|}^2|z_2{|}^2$
$\Rightarrow |z_1{|}^2-|z_1{|}^2|z_2{|}^2+4|z_2{|}^2-4=0$
$\Rightarrow |z_1{|}^2(1-|z_2{|}^2)+4(|z_2{|}^2-1)=0$
$\Rightarrow (|z_2{|}^2-1)(|z_1{|}^2-4)=0$
$\Rightarrow |z_1|=2$ as $|z_2|\ne 1$

Alternate :
Put $z_2=0$ in $\left|\frac{\overline{z_1}-2\overline{z_2}}{2-z_1\overline{z_2}}\right|=1,$ we get
$\left|\frac{\overline{z_1}}{2}\right|=1$
$\Rightarrow |z_1|=2$ as $|z_1|=|\overline{z_1}|$