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  4. Let z ∈ C be such that |z| < 1. If ω = (5 + 3z/5( 1 - z)), then :

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

JEE MainJEE Main 2019Complex Numbers and Quadratic Equations

Solution:

$\left|z\right| < 1 $
$ 5\omega\left(1-z\right)=5+3z $
$ 5\omega - 5\omega z = 5 +3z $
$ z = \frac{5\omega-5}{3+5\omega}$
$ \left|z \right| = 5 \left|\frac{\omega-1}{3+5\omega} \right|<1 $
$ 5 \left|\omega-1\right|< \left|3+5\omega \right| $
$ 5 \left|\omega-1\right| < 5 \left|\omega+ \frac{3}{5}\right| $
$ \left|\omega-1\right| < 5 \left|\omega - \left( - \frac{3}{5}\right)\right| $

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