I'm struggling to reconcile that a complex integrable function may be multi-branched with the statement that its integral is contour-independent. Consider f(z) = z^(1/n), n natural, n-branched and its integral from z_0 to z_1.(adsbygoogle = window.adsbygoogle || []).push({});

On the way from z_0 to z_1 I can take a few detours near the origin where in each evaluation of z^(1/n) I can choose a branch. So I may pass across, say, the real axes at z=1 and choose, for n=2, -1 while I'm still below it and 1 as soon as I have crossed it. On the way back, however, I stick with one branch.

That wouldn't work. So my question is, what is the mathematical correct statement then? It obviously can't be as broad as "the integral of complex integrable function is contour independent".

Should we say "The integral of a complex integrable function is contour independent if its evaluated so that it's differentiable along the path" (thus forbidding any "switch of branch)? What if we consider more esoteric, multi-branched functions where two different branches concide at a point (which would then still allow us to jump from one onto the other).

Or should we say "Branches can always be choosen such that the integral from z_0 to z_1 equals any other integral from z_0 to z_1"?

On a second thought: Perhaps the "function" is defined by an appropriate choice of a branch at each point a-priori. That would clearly make sense. We require the function to be holomorphic, which it only is under a certain choice of branching.

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# Complex (contour indep.) integral of a branched integrand

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