Question

Let $z_1$ and $z_2$ be two complex numbers such that $\left|\frac{z_1-2z_2}{2-z_1\overline{z_2}}\right|=1$ and $|z_2|\ne 1$. Then the value of $|z_1|$ is

• A. $2$
• B. $4$
• C. $1$
• D. $8$

Answer 1

The correct option is A $2$

$\left|\frac{z_1-2z_2}{2-z_1\overline{z_2}}\right|=1\Rightarrow |z_1-2z_2|=|2-z_1\overline{z_2}|$
Squaring on both sides
$|z_1-2z_2{|}^2=|2-z_1\overline{z_2}{|}^2$
$\Rightarrow (z_1-2z_2)(\overline{z_1}-2\overline{z_2})=(2-z_1\overline{z_2})(2-\overline{z_1}{z}_2)$
$\Rightarrow z_1\overline{z_1}-2z_1\overline{z_2}-2\overline{z_1}{z}_2+4z_2\overline{z_2}$
$=4-2z_1\overline{z_2}-2\overline{z_1}{z}_2+z_1\overline{z_1}{z}_2\overline{z_2}\Rightarrow |z_1{|}^2-|z_1{|}^2|z_2{|}^2+4|z_2{|}^2-4=0\Rightarrow (|z_1{|}^2-4)(1-|z_2{|}^2)=0\Rightarrow |z_1{|}^2=4(\because |z_2|\ne 1)$

Hence, $|z_1|=2$

Alternate solution:
Put $z_2=0$ (or any other specific value whose modulus is not $1$)
$\left|\frac{z_1}{2}\right|=1\Rightarrow |z_1|=2$