If $z_{1} \& z_{2}$ are two complex numbers such that $\left|z_{1}\right|

Question

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

Professor PC Thomas Classes · Answer

To prove : $\left|1-z_{1} \bar{z}_{2}ight|<\left|z_{1}-z_{2}ight|$\ $\Rightarrow\left(1-z_{1} \bar{z}_{2}ight)\left(1-\bar{z}_{1} z_{2}ight)<\left(z_{1}-z_{2}ight)\left(\bar{z}_{1}-\bar{z}_{2}ight)$\ $\Rightarrow\left(1-z_{1} \bar{z}_{2}-\bar{z}_{1} z_{2}+\left|z_{1}ight|^{2}\left|z_{2}ight|^{2}ight)$\ $<\left(\left|z_{1}ight|^{2}-z_{1} \bar{z}_{2}-z_{2} \bar{z}_{1}+\left|z_{2}ight|^{2}ight)$\ $\Rightarrow\left(1-\left|z_{1}ight|^{2}ight)-\left|z_{2}ight|^{2}\left(1-\left|z_{1}ight|^{2}ight)<0$\ $\Rightarrow\left(1-\left|z_{2}ight|^{2}ight)\left(1-\left|z_{1}ight|^{2}ight)<0$,\ Which is obvious as $\left|z_{1}ight|<1<\left|z_{2}ight|$\