I am confused how the existence of a complex anti derivative on an open set implies the path integral property. The proof of this is shown here and it is very simple.
Is this not a trivial counter example?
U={z∈C:|z|>0.5 ∧|z|<1.5}
g(z):=1/z
I believe g(z) has an anti-derivative on all of U but ∫γg(z)dz=0 on γ of a unit circle.
This would make more sense if the requirement was U is simply connected, but that requirement does not seem to be used in the proof. What am I missing?
Answer
We can define G(z)=Arg(z) on all of U but it will be discontinuous, and therefore its derivative is not defined everywhere. Then G is not an antiderivative of g on all of U.
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