Final answer:
By using a contour integral and the Residue Theorem, we show that the function f(z) = 2z/(z-1) does not have an antiderivative on the domain C because the integral around a closed path that encloses the singularity is not zero.
Step-by-step explanation:
To demonstrate that the function f(z) = 2z/(z-1) does not have an antiderivative on the domain C, we can use a contour integral around a closed path that encircles the point z = 1, where the function has a singularity (a point of discontinuity).
Consider the contour integral of f(z) around a simple closed curve γ that encloses the singular point z = 1. The integral of f(z) over this curve is given by:
∮ γ f(z) dz = ∮ γ • 2z/(z-1) dz
Since the function has a pole of order 1 at z = 1, by the Residue Theorem, the integral equals 2πi times the residue of f(z) at z = 1, which is not zero. This shows that f(z) cannot have an antiderivative in the entire complex plane because the integral around a closed path is not zero as it would be if f(z) had an antiderivative (according to the Fundamental Theorem of Calculus for complex functions).
This non-zero integral around a simple closed loop implies that f(z) does not satisfy the necessary condition for the existence of an antiderivative – that is, the integral of f(z) over any closed contour must be zero. Hence, f(z) does not have an antiderivative on the domain C.