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

|z| < 1
5ω(1 - z) = 5 + 3z
5ω - 5ωz = 5 + 3z
z(3 + 5ω) = 5ω - 5
z = (5ω - 5)/(3 + 5ω)
|z| = |(5ω - 5)/(3 + 5ω)| < 1
|5ω - 5| < |3 + 5ω|
|ω - 1| < |3/5 + ω|
Let ω = x + iy
|(x - 1) + iy| < |(3/5 + x) + iy|
(x - 1)² + y² < (3/5 + x)² + y²
(x - 1)² < (3/5 + x)²
x² - 2x + 1 < x² + 6x/5 + 9/25
1 - 2x < 6x/5 + 9/25
25 - 50x < 30x + 9
25 - 9 < 80x
16 < 80x
x > 16/80 = 1/5
Re(ω) > 1/5
5Re(ω) > 1