Question

A complex number $z$ is the said to be unimodular if $|z|=1 .$ Suppose $z_{1}$ and $z_{2}$ are complex number such that $\frac{z_{1}-2 z_{2}}{2-z_{1} z_{2}}$ is unimodular and $z _{2}$ is not unimodular. Then the point $z_{1}$ lies on a :

• A. Straight line parallel to x-axis
• B. Straight line parallel to y-axis
• C. Circle of radius 2
• D. Circle of radius $\sqrt{2}$

Answer 1

The Correct Option is C

$\left|\frac{z_{1}-2 z_{2}}{2-z_{1} \bar{z}_{2}}\right|=1$
$\left(z_{1}-2 z_{2}\right)\left(\bar{z}_{1}-2 \bar{z}_{2}\right)=\left(2-z_{1} \bar{z}_{2}\right)\left(2-\bar{z}_{1} z_{2}\right)$
$\left|z_{1}\right|^{2}-2 z_{1} \bar{z}_{2}-2 z_{2} \bar{z}_{1}+4\left|z_{2}\right|^{2} $
$=4-2 \bar{z}_{1} z_{2}-2 z_{1} \bar{z}_{2} +\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}$
$\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}-\left|z_{1}\right|^{2}-4\left|z_{2}\right|^{2}+4=0$
$\left(\left|z_{1}\right|^{2}-4\right)\left(\left|z_{2}\right|^{2}-1\right)=0$
$\Rightarrow\left|z_{1}\right|=2$