Question

If | $\frac{z_1-2z_2}{2-z_1\overline{z_2}}$| = 1, |$z_2$| $\ne$ 1, then |$z_1$| =

• A. 4
• B. 2
• C. 1
• D. None
  

Answer 1

The correct option is B

2

| $\frac{z_1-2z_2}{2-z_1\overline{z_2}}$| = 1 $\Rightarrow$ $|z_1-2z_2|=|2-z_1\overline{z_2}|$

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

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

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

$\Rightarrow$ $|z_1{|}^2$ = 2 as $|z_2{|}^2$ $\ne$ 1