Let z

Question:

Let $z \in \mathrm{C}$ be such that $|z|<1$. If $\omega=\frac{5+3 z}{5(1-z)}$, then :

  1. (1) $5 \operatorname{Re}(\omega)>4$

  2. (2) $4 \operatorname{Im}(\omega)>5$

  3. (3) $5 \operatorname{Re}(\omega)>1$

  4. (4) $5 \operatorname{Im}(\omega)<1$


Correct Option: , 3

Solution:

$\omega=\frac{5+3 z}{5-5 z} \Rightarrow 5 \omega-5 \omega z=5+3 z$

$\Rightarrow 5 \omega-5=z(3+5 \omega) \Rightarrow z=\frac{5(\omega-1)}{3+5 \omega}$

$\because|z|<1, \therefore 5|\omega-1|<|3+5 \omega|$

$\Rightarrow 25(\omega \bar{\omega}-\omega-\bar{\omega}+1)<9+25 \omega \bar{\omega}+15 \omega+15 \bar{\omega}$

$\left(\because|z|^{2}=z \bar{z}\right)$

$\Rightarrow 16<40 \omega+40 \bar{\omega} \Rightarrow \omega+\bar{\omega}>\frac{2}{5} \Rightarrow 2 \operatorname{Re}(\omega)>\frac{2}{5}$

$\Rightarrow \operatorname{Re}(\omega)>\frac{1}{5}$

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