Question

State true or false:

Let $\left|\right({\overline{z}}_1-2{\overline{z}}_2\left)/\right(2-z_1{\overline{z}}_2\left)\right|=1$ and $\left|z_2\right|\ne 1$, where $z_1$ and $z_2$ are complex numbers, then $\left|z_1\right|=2$.

Answer 1

$\left|\frac{{\overline{z}}_1-2{\overline{z}}_2}{2-z_1{\overline{z}}_2}\right|=1$
or ${|{\overline{z}}_1-2{\overline{z}}_2|}^2={|2-z_1{\overline{z}}_2|}^2$
or $({\overline{z}}_1-2{\overline{z}}_2)\left(\overline{{\overline{z}}_1-2{\overline{z}}_2}\right)=(2-z_1{\overline{z}}_2)\left(\overline{2-z_1{\overline{z}}_2}\right)$
or $({\overline{z}}_1-2{\overline{z}}_2)({\overline{z}}_1-2{\overline{z}}_2)=(2-z_1{\overline{z}}_2)(2-z_1{\overline{z}}_2)$
or $z_1{\overline{z}}_1-2{\overline{z}}_1{z}_2-2z_1{\overline{z}}_2+4z_2{\overline{z}}_2=4-2{\overline{z}}_1{z}_2-2z_1{\overline{z}}_2+z_1{\overline{z}}_1{z}_2{\overline{z}}_2$
or ${\left|z_1\right|}^2+4{\left|z_2\right|}^2=4+{\left|Z_1\right|}^2{\left|z_2\right|}^2$
or ${\left|z_1\right|}^2-{\left|z_1\right|}^2{\left|z_2\right|}^2+4{\left|z^2\right|}^2-4=0$
or ${\left|z_1\right|}^2-(1-{\left|z_2\right|}^2)+4({\left|z_2\right|}^2-1)=0$
or ${\left|z_2\right|}^2-1\left)\right({\left|z_1\right|}^2-4)=0$
or $\left|z_1\right|=2$
(as $\left|z_2\right|\ne 1$)