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

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

$⇒ 5\omega\left(1 - z\right) = 5 + 3z$

$⇒ 5\omega - 5\omegaz = 5 + 3z$

$⇒ z = \dfrac{5\omega - 5}{3 + 5\omega}$

Using the given condition \left|z\right| = 5\left|\dfrac{\omega - 1}{3 + 5\omega}\right| < 1$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/6d93483a-6d23-4e29-a9d9-139fafe1e404-image8.png" style="width: 250px; height: 160px;" /><br />$⇒ 5\left|\omega - 1\right| < \left|3 + 5\omega\right|$</p><p>$⇒ 5\left|\omega - 1\right| < 5\left|\omega + \dfrac{3}{5}\right|$</p><p>$⇒ \left|\omega - 1\right| < \left|\omega - \left(- \dfrac{3}{5}\right)\right|$</p><p>We know that the locus of a complex number${z}_{1}$satisfying \left|{z}_{1} - a\right| = \left|{z}_{2} - b\right|$ is the perpendicular bisector of the line segment joining the points $a$ and $b .$

Hence, the locus of the point $\omega$ satisfying \left|\omega - 1\right| = $\left|\omega - \left(- \dfrac{3}{5}\right)\right|$ is the line $x = \dfrac{1 - \dfrac{3}{5}}{2} = \dfrac{1}{5}$

Hence, $Re\left(\omega\right) > \dfrac{1}{5}$

$⇒ 5Re\left(\omega\right) > 1 .$