Question

$z_{1}$ and $z_{2}$ are two complex numbers such that $\frac{z_{1}-2 z_{2}}{2-z_{1} \bar{z}_{2}}$ is unimodular whereas $z_{2}$ is not unimodular. Then $\left|z_{1}\right|=$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Clearly $\left|\frac{z_{1}-2 z_{2}}{2-z_{1} \bar{z}_{2}}\right|=1$
$\Rightarrow \left(\frac{z_{1}-2 z_{2}}{2-z_{1} \bar{z}_{2}}\right)\left(\frac{\bar{z}_{1}-2 \bar{z}_{2}}{2-\bar{z}_{1} z_{2}}\right)=1$
$\Rightarrow z_{1} \bar{z}_{1}-2 z_{1} \bar{z}_{2}-2 \bar{z}_{1} z_{2}+4 z_{2} \bar{z}_{2}$
$=4-2 \bar{z}_{1} z_{2}-2 z_{1} \bar{z}_{2}+z_{1} z_{2} \bar{z}_{1} \bar{z}_{2}$
$\Rightarrow \left|z_{1}\right|^{2}+4\left|z_{2}\right|^{2}=4+\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}$
$\Rightarrow \left|z_{1}\right|^{2}\left[1-\left|z_{2}\right|^{2}\right]=4\left[1-\left|z_{2}\right|^{2}\right]$
$\Rightarrow \left|z_{1}\right|^{2}=4$
$\left(\because\left|z_{2}\right| \neq 1\right)$
$\Rightarrow \left|z_{1}\right|=2$