Question

State whether the following statement is true or false.

Let $z_1,z_2$ be two complex numbers such that $\frac{z_1-{2z}_2}{2-z_2{\overline{z}}_2}$ is unimodular. If $z_2$ is not unimodular then $\left|z_1\right|=2$.

• A. True
• B. False

Answer 1

The correct option is A True

Step-by-step explanation:

if $z$ unimodular then |z|=1 also use property of moduls i.e $z\overline{z}=|z{|}^2$

given $z^2$ is not unimodular $|z{|}^2\ne 1$

and $\frac{z_1-2z_2}{2-z_1{\overline{z}}_2}$ is unimodular

$\frac{z_1-2z_2}{2-z_1{\overline{z}}_2}=1$

⇒$|z_1-2z_2{|}^2=|2-z_1{\overline{z}}_2{|}^2$

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

$\because z\overline{z}=|z{|}^2$

$\Rightarrow |z_2{|}^2+4|z_2{|}^2-2{\overline{z}}_1{z}_2-2z_1{\overline{z}}_2=0$

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

$|z_2|\ne 1$

$z_1=x+iy$

$x^2+y^2=(2{)}^2$

$z_1$ lies on radius is $2$.

Hence $|z_1|=2$.