This is what Wolfram/Alpha computing engine says about your problem:
"inequalities are not well-defined in the complex plane (the field of complex numbers is not totally-ordered)"
+33661 If we set z = a + bi, where a and b are Real, then the blue surface in the 3d graph below shows where the valid parameters lie.
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+33661 By setting (a^2 + b^2 - a - b) = 2(a^2 + b^2 - 2b + 1) we find that the boundary within which Re((z-1)/(z-i))>=2 is given by
(a + 1/2)^2 + (b - 3)^2 = r^2 where r^2 = 33/4 This is a circle of radius ≈ 2.872, centred on (-1/2, 3).
Any combination of a and b such that (a + 1/2)^2 + (b - 3)^2 <= r^2 produces a value of z that satisfies Re((z-1)/(z-i))>=2
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