Friday, 9 March 2018

Complex anti-derivative and the path integral property



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={zC:|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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