|z - 3| |-----| >= 1 |z - 2|
Abs((z - 3)/(z - 2)) >= 1
x0 = 0
|-3 + z| |------| >= 1 |-2 + z|
(-oo, 2) U (2, 5/2]
x in Union(Interval.open(-oo, 2), Interval.Lopen(2, 5/2))
Or(And(z <= 5/2, 2 < z), And(-oo < z, z < 2))
((z <= 5/2)∧(2 < z))∨((-oo < z)∧(z < 2))