Question

Q. $z_1$and $z_2$ are two complex numbers such that $\frac{z_1-2z_2}{2-z_1\overline{z_2}}$ is unimodular whereas $z_2$ is not a ​unimodular.
Then |$z_1$| is

1. 1
2. 2
3. 3
4. 4

Answer 1

Dear student
$$\begin{gathered} \mathrm{A}\mathrm{complex}\mathrm{number}\mathrm{z}\mathrm{is}\mathrm{said}\mathrm{to}\mathrm{be}\mathrm{unimodular}\mathrm{if}\left|\mathrm{z}\right|=1. \\ \mathrm{Consider},\left|\frac{{\mathrm{z}}_1-2{\mathrm{z}}_2}{2-{\mathrm{z}}_1\overline{{\mathrm{z}}_2}}\right| \\ \mathrm{Since}\left|\frac{{\mathrm{z}}_1-2{\mathrm{z}}_2}{2-{\mathrm{z}}_1\overline{{\mathrm{z}}_2}}\right|\mathrm{is}\mathrm{unimodular}\mathrm{then}\left|\frac{{\mathrm{z}}_1-2{\mathrm{z}}_2}{2-{\mathrm{z}}_1\overline{{\mathrm{z}}_2}}\right|=1 \\ \Rightarrow |{\mathrm{z}}_1-2{\mathrm{z}}_2|=|2-{\mathrm{z}}_1\overline{{\mathrm{z}}_2}| \\ \Rightarrow |{\mathrm{z}}_1-2{\mathrm{z}}_2{|}^2=|2-{\mathrm{z}}_1\overline{{\mathrm{z}}_2}{|}^2 \\ \Rightarrow \left({\mathrm{z}}_1-2{\mathrm{z}}_2\right)\left(\overline{{\mathrm{z}}_1}-2\overline{{\mathrm{z}}_2}\right)=\left(2-{\mathrm{z}}_1\overline{{\mathrm{z}}_2}\right)\left(2-\overline{{\mathrm{z}}_1}{\mathrm{z}}_2\right)\left[\mathrm{using}{\left|\mathrm{z}\right|}^2=\mathrm{z}\overline{\mathrm{z}}\right] \\ \Rightarrow {\mathrm{z}}_1\overline{{\mathrm{z}}_1}-\cancel{2{\mathrm{z}}_1\overline{{\mathrm{z}}_2}-2{\mathrm{z}}_2\overline{{\mathrm{z}}_1}}+4{\mathrm{z}}_2\overline{{\mathrm{z}}_2}=4-\cancel{2\overline{{\mathrm{z}}_1}{\mathrm{z}}_2-2{\mathrm{z}}_1\overline{{\mathrm{z}}_2}}+{\mathrm{z}}_1\overline{{\mathrm{z}}_1}{\mathrm{z}}_2\overline{{\mathrm{z}}_2} \\ \Rightarrow {\left|{\mathrm{z}}_1\right|}^2+4{\left|{\mathrm{z}}_2\right|}^2=4+{\left|{\mathrm{z}}_1\right|}^2{\left|{\mathrm{z}}_2\right|}^2 \\ \Rightarrow {\left|{\mathrm{z}}_1\right|}^2+4{\left|{\mathrm{z}}_2\right|}^2-4-{\left|{\mathrm{z}}_1\right|}^2{\left|{\mathrm{z}}_2\right|}^2=0 \\ \Rightarrow {\left|{\mathrm{z}}_1\right|}^2\left(1-{\left|{\mathrm{z}}_2\right|}^2\right)-4\left(1-{\left|{\mathrm{z}}_2\right|}^2\right)=0 \\ \Rightarrow \left(1-{\left|{\mathrm{z}}_2\right|}^2\right)\left({\left|{\mathrm{z}}_1\right|}^2-4\right)=0 \\ \mathrm{Since}\left|{\mathrm{z}}_2\right|\ne 1 \\ \Rightarrow {\left|{\mathrm{z}}_1\right|}^2-4=0 \\ \Rightarrow {\left|{\mathrm{z}}_1\right|}^2=4 \\ \Rightarrow \left|{\mathrm{z}}_1\right|=2 \end{gathered}$$
Regards