Question

Let $|\frac{z_1-2z_2}{2-z_1\overline{z_2}}|=1$ and $|z_2|\ne 1,$ where $z_{1}and{z}_2$ are complex numbers. Find the value of $|z_1|.$

Answer 1

$\left|\frac{z_1-2z_2}{2-z_1\cdot {\overline{z}}_2}\right|=1$

$\Rightarrow (z_1-2z_2)({\overline{z}}_1-2{\overline{z}}_2=(2-z_1\cdot \overline{z}\left)\right(2-{\overline{z}}_1\cdot z_2)$

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

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

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

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

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

$\Rightarrow |z_1{|}^2-4=0$ $\because z_2\ne 1$

$\Rightarrow z_1=2$