Lack of existence of antiderivative for complex function I have been asked to show that the functio

Marissa Singh

Marissa Singh

Answered question

2022-04-07

Lack of existence of antiderivative for complex function
I have been asked to show that the function f ( z ) = z z 2 + 1 does not have an antiderivative on the subset { z C : 1 < | z | }. However I don't see why this is the case, we can write z z 2 + 1 = 1 2 ( 1 z + i + 1 z i ) whose naive antiderivative (using ln) has branch points at so taking a straight line between these points as our branch cut should allow us to make it continuous and single valued on our region. What am I missing?
I know I am making a mistake somewhere as the residue theorem tells me the integral around a circle of radius 2 has value 2 π i and not 0 so no antiderivative exists. Any help is much appreciated.

Answer & Explanation

charringpq49u

charringpq49u

Beginner2022-04-08Added 23 answers

Explanation:
Consider a positively oriented circle Γ of radius > 1 centred at the origin.
1 2 π i Γ z z 2 + 1 d z
is the sum of the residues of z / ( z 2 + 1 ), namely 1. But if your function had an antiderivative on that region, the result would have to be 0.

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