Residues and existence of complex antiderivative I know that there already is a similar question ou

Pattab 2022-07-06 Answered
Residues and existence of complex antiderivative
I know that there already is a similar question out there but I am not really satisfied with that answer. I think it is too complicated. Assume that G C is open and simply connected, P C has finite cardinality and f : G P C is holomorphic. Then
f has an antiderivative on its domain R e s p ( f ) = 0   p P.
Can I argue the following way? The crucial part is the second implication.
: Since f has an antiderivative, all integrals over closed rectifiable curves vanish. If I let p P be arbitrary and ε > 0 so small that B ε ( p ) G P, then according to the residue theorem:
0 = B ε ( p ) f ( z )   d z = 2 π i R e s p ( f )
: For any closed rectifiable curve γ we have:
γ f ( z )   d z = p P I p ( γ ) R e s p ( f ) = 0
Therefore the existence of a primitive.
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Answers (1)

Alexia Hart
Answered 2022-07-07 Author has 19 answers
Step 1
Under the given conditions on G, P, and f, the following three statements are equivalent:
(a) f has an antiderivative in G.
(b) Res p ( f ) = 0 for all p P.
(c) γ f ( z ) d z = 0 for all closed rectifiable curves γ in G.
Step 2
What you have demonstrated is that ( a ) ( b ) and that ( b ) ( c ).
It remains to show that ( c ) ( a ), and that can be done by showing that F ( z ) = z 0 z f ( w ) d w (which is well-defined under the condition (c)) satisfies F = f
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