Existence of an antiderivative in U ∪ VI am trying to prove the following:Let U , V ⊂ C be open and connected sets such that U ∩ V is connected as well.Let f be a holomorphic funtion with complex antiderivative on Uand V.Then f has a complex antiderivative on U ∪ V.Let F 1 be the antiderivative on U and F 2 be the antiderivative on V then F 1 ′ = F 2 ′ on the conncted set U ∩ V.Hence F 1 = F 2 + c.I dont know how to get to get the antiderivative on U ∪ V from here.

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Existence of an antiderivative in   U  ∪  VI am trying to prove the following:Let   U  ,  V  ⊂      C   be open and connected sets such that   U  ∩  V is connected as well.Let f be a holomorphic funtion with complex antiderivative on Uand V.Then f has a complex antiderivative on   U  ∪  V.Let       F    1   be the antiderivative on U and       F    2   be the antiderivative on V then       F    1                          ′              =      F    2                          ′             on the conncted set   U  ∩  V.Hence       F    1    =      F    2    +  c.I dont know how to get to get the antiderivative on   U  ∪  V from here.

pheniankang · Accepted Answer

Explanation:Here&#039;s the idea. If you knew that       F    1    =      F    2   on the intersection, then you could define an antiderivative F on   U  ∪  V by just saying   F  (  z  )  =      F    1    (  z  ) for   z  ∈  U and   F  (  z  )  =      F    2    (  z  ) for   z  ∈  V. This is well-defined because       F    1    (  z  )  =      F    2    (  z  ) if   z  ∈  U  ∩  V.Unfortunately, you might not have       F    1    =      F    2   on   U  ∩  V. But can you see a way you could modify       F    2   such that this is true?

Existence of an antiderivative in U ∪ V I am trying to prove the following: Let

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