Antiderivatives in complex analysisLet U be an open set in C and f : U → E be holomorphic on U. Let now γ be a path in U that begins with z 1 ∈ U and ends with z 2 ∈ U .. I wonder for which assumptions the following formula ∫ γ d f d z ( z ) = f ( z 2 ) − f ( z 1 ) is true?I have found many counterexamples (such as f ′ ( z ) = 1 z ) where the above formula fails if the set U is not simply connected. Here is a related result.Another question, if f is well defined on U ¯ and z 1 ∈ ∂ U is the above formula also true?

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Antiderivatives in complex analysisLet U be an open set in       C   and   f  :  U  →  E be holomorphic on U. Let now   γ be a path in U that begins with       z    1    ∈  U and ends with       z    2    ∈  U  .. I wonder for which assumptions the following formula       ∫    γ              d      f              d      z        (  z  )  =  f  (      z    2    )  −  f  (      z    1    ) is true?I have found many counterexamples (such as       f    ′    (  z  )  =      1    z  ) where the above formula fails if the set U is not simply connected. Here is a related result.Another question, if f is well defined on             U      ¯       and       z    1    ∈  ∂  U is the above formula also true?

Tyler Velasquez · Accepted Answer

Step 1If f is holomorphic in   U  ⊂      C   and   γ is a (piecewise differentiable) path in U joining       z    1   and       z    2   then      ∫    γ        f    ′    (  z  )    d  z  =      ∫    0    1        f    ′    (  γ  (  t  )  )      γ    ′    (  t  )    d  t  =            [        f  (  γ  (  t  )  )                    ]                    t      =      0              t      =      1        =  f  (      z    2    )  −  f  (      z    1    )Step 2So that relation holds always, even in multiply connected domains.      ∫                  |            z              |            =      1            1    z      d  z  =  2  π  i  ≠  0 is not a counter-example, because 1/z is not the derivative of a holomorphic function f in any domain containing the unit circle.

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