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

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

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

$\Rightarrow 5\omega -5\omega z=5+3z$

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

Using the given condition $\left|z\right|=5\left|\frac{\omega -1}{3+5\omega }\right|<1$

$\Rightarrow 5\left|\omega -1\right|<\left|3+5\omega \right|$

$\Rightarrow 5\left|\omega -1\right|<5\left|\omega +\frac{3}{5}\right|$

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

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 $\left|\omega -1\right|=\left|\omega -\left(-\frac{3}{5}\right)\right|$ is the line $x=\frac{1-{\displaystyle \frac{3}{5}}}{2}=\frac{1}{5}$

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

$\Rightarrow 5Re\left(\omega \right)>1.$