Complex antiderivativeI 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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Complex antiderivativeI 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?

SweallySnicles3 · Accepted Answer

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&#039;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&#039;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&#039;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

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

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New Questions in Calculus 1