Question

If $z_{1}$ and $z_{2}$ are two complex numbers such that $\left|z_{1}\right|<1<\left|z_{2}\right|$ then $\left|\frac{1-z_{1} \bar{z}_{2}}{z_{1}-z_{2}}\right|$

• A. equal to 1
• B. greater than 1
• C. less than 1
• D. none of these

Answer 1

Answer: (C)

Given that $\left|z_{1}\right|<1<\left|z_{2}\right|$
Let $\left|\frac{1-z_{1} \bar{z}_{2}}{z_{1}-z_{2}}\right|<1$
$\Rightarrow\left|1-z_{1} \bar{z}_{2}\right|<\left|z_{1}-z_{2}\right|$
$\Rightarrow\left|1-z_{1} \bar{z}_{2}\right|^{2}<\left|z_{1}-z_{2}\right|^{2}$
$\Rightarrow \left(1-z_{1} \bar{z}_{2}\right)\left(\overline{1-z_{1} \bar{z}_{2}}\right)<\left(z_{1}-z_{2}\right)\left(\overline{z_{1}-z_{2}}\right)$
$\Rightarrow \left(1-z_{1} \bar{z}_{2}\right)\left(1-\bar{z}_{1} z_{2}\right)<\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)$
$\Rightarrow 1-z_{1} \bar{z}_{2}-\bar{z}_{1} z_{2}+z_{1} \bar{z}_{1} z_{2} \bar{z}_{2}< z_{1} \bar{z}_{1}-z_{1} \bar{z}_{2}-\bar{z}_{1} z_{2}+z_{2} \bar{z}_{2}$
$\Rightarrow 1+\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}< \left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}$
$\Rightarrow \left(1-\left|z_{1}\right|^{2}\right)\left(1-\left|z_{2}\right|^{2}\right)<0$
which is obviously true as $\left|z_{1}\right|<1<\left|z_{2}\right|$ and hence, $\left|z_{1}\right|^{2}<1$ $<\left|z_{2}\right|^{2}$.