Complex antiderivative I am confused on a couple things: 1.) Why is it that an integral of a comp

kolutastmr 2022-07-02 Answered
Complex antiderivative
I am confused on a couple things:
1.) Why is it that an integral of a complex valued function of a complex variable exists if f(z(t)) is piecewise continuous (and/or piecewise continuous on C ) and not continuous, like a real?
2.) Why is it that one cannot make use of an antiderivative to evaluate an integral of a function like 1/z, on a contour of something like z = 2 e i θ positively oriented with π θ π? That is, because F ( z ) = 1 / z is undefined at 0, it is disqualified (I think). But why?
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Answers (1)

SweallySnicles3
Answered 2022-07-03 Author has 21 answers
Explanation:
If the function f(z) has an antiderivative F(z) on a domain that includes a contour Γ that goes from point a to point b, then Γ f ( z )   d z = F ( b ) F ( a ). The problem is that in many cases such an antiderivative can't exist. A case in point is your example, Γ d z z , where Γ is the counterclockwise circle of radius 2 centred at 0. You'd like to say that an antiderivative of f ( z ) = 1 / z is log ( z ), but there is no version of log(z) that is continuous on that whole circle. Indeed that must be so, because since b = a (this contour ends in the same place where it starts), you'd have Γ d z z = F ( a ) F ( a ) = 0. But, parametrizing the contour as z ( t ) = 2 e i t , Γ d z z = π π 2 i e i t   d t 2 e i t = π π i   d t = 2 π i

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