Question:
Let $z \in \mathrm{C}$ be such that $|z|<1$. If $\omega=\frac{5+3 z}{5(1-z)}$, then :
Correct Option: , 3
Solution:
$\omega=\frac{5+3 z}{5-5 z} \Rightarrow 5 \omega-5 \omega z=5+3 z$
$\Rightarrow 5 \omega-5=z(3+5 \omega) \Rightarrow z=\frac{5(\omega-1)}{3+5 \omega}$
$\because|z|<1, \therefore 5|\omega-1|<|3+5 \omega|$
$\Rightarrow 25(\omega \bar{\omega}-\omega-\bar{\omega}+1)<9+25 \omega \bar{\omega}+15 \omega+15 \bar{\omega}$
$\left(\because|z|^{2}=z \bar{z}\right)$
$\Rightarrow 16<40 \omega+40 \bar{\omega} \Rightarrow \omega+\bar{\omega}>\frac{2}{5} \Rightarrow 2 \operatorname{Re}(\omega)>\frac{2}{5}$
$\Rightarrow \operatorname{Re}(\omega)>\frac{1}{5}$