Question

If $|z_1|<1\text{and}|\frac{z_1-z_2}{1-{\overline{z}}_1{z}_2}|<1,\text{then}|z_2|>1$

• A. True
• B. False

Answer 1

The correct option is B False

Given.

$|z_i|<1$

$\left|\frac{z_1-z_2}{1-z_1{z}_2}\right|<1$

$|z_1-z_2|<|1-{\overline{z}}_1{z}_2|$

${|z_1-z_2|}^2<{|1-{\overline{z}}_1{z}_2|}^2$

$(z_1-z_2)({\overline{z}}_1-{\overline{z}}_2)<(1-{\overline{z}}_1{z}_2)(1-z_1{\overline{z}}_2)$

$z_1{\overline{z}}_1-z_2{\overline{z}}_1-{\overline{z}}_2{z}_1+z_2{\overline{z}}_2<1-z_1{\overline{z}}_2-{\overline{z}}_1{z}_2+\left(z_1{\overline{z}}_1\right)\left(z_2{\overline{z}}_2\right)$

${\left|z_1\right|}^2+{\left|z_2\right|}^2⟨{1+|z_1|}^2{\left|z_2\right|}^2$

${\left|z_1\right|}^2+{|z_2|}^2+{\left|z_1\right|}^2{\left|z_2\right|}^2-1<0$

${\left|z_2\right|}^2\left(\right(1-{\left|z_1\right|}^2)-1{(1-z_1)}^2)<0$

$({|z_2|}^2-1)(1-{|z_1|}^2)<0$

$(|z_2{|}^2-1)(|z_2{|}^2-1)>0$

$|z_1{|}^2-1<0$

$\Rightarrow |{z_2|}^2-1<0$

${\left|z_2\right|}^2<1$

$\left|z_2\right|<1$

option $B$ is Correct