Cauchy integral theorem states that the contour integration of a complex harmonic function along a closed simply connected path=0. What if this simply connected path is drawn over a Riemann surface of function like ##f(z)=\sqrt z##. Will that be possible in the first place? and will the integration over that path yield zero?(adsbygoogle = window.adsbygoogle || []).push({});

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# I Contour integration over Riemann surface

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