If $| z _ { 1 } | < 1 \text { and } | \frac { z _ { 1 } - z _ { 2 } } { 1 - \overline { z } _ { 1 } z _ { 2 } } | < 1 , \text { then } | z _ { 2 } | > 1$

Indeterminate

Given.

$|z _i|<1$

$\left|\frac{z_{1}-z_{2}}{1-z_{1} z_{2}}\right|<1 \\$

$\left|z_{1}-z_{2}\right|<\left|1-\bar{z}_{1} z_{2}\right| \\$

$\left|z_{1}-z_{2}\right|^{2}<\left|1-\bar{z}_{1} z_{2}\right|^{2}$

$\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)<\left(1-\bar{z}_{1} z_{2}\right)\left(1-z_{1} \bar{z}_{2}\right)$

$z_{1} \bar{z}_{1}-z_{2} \bar{z}_{1}-\bar{z}_{2} z_{1}+z_{2} \bar{z}_{2} \quad<1-z_{1} \bar{z}_{2}-\bar{z}_{1} z_{2}+\left(z_{1} \bar{z}_{1}\right)\left(z_{2} \bar{z}_{2}\right)$

$\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2} \quad\left\langle\left.\quad1+| z_{1}\right|^{2}\left|z_{2}\right|^{2}\right.$

$\left|z_{1}\right|^{2}+\left.|z_{2}\right|^{2}+\left|z_{1}\right|^{2}\left|z_{2}\right|^{2}-1<0$

$\left.\left|z_{2}\right|^{2}\left((1-\left|z_{1}\right|^{2}\right)-1\left(1-z_{1}\right)^{2}\right)< 0$

$\left(\left.| z_2\right|^{2}-1\right)\left(1-\left.|z_{1}\right|^{2}\right)<0$

$\left.\left.(|z_2|^{2}-1)\right(|z_2|^{2}-1\right)>0$

$|z_1|^{2}-1<0$

$\Rightarrow\left|\left.z_{2}\right|^{2}-1<0\right.$

$\left|z_{2}\right|^{2}<1$

$\left|z_{2}\right|<1$

option $B$ is Correct