Let $D$ be a domain, say a rectangle in the complex plane, with three distinct holes inside. Now let $f$ be analytic on $D$ and on the boundary of $D$. My question is why is the contour integral of $f$ along the outer boundary equal to the sum of the contour integrals along the three "inner" boundaries. I think this is true because you can continuously deform the outer boundary (the contour) so that it's the three inner circles connected by line segments. And the claim follows by the deformation invariance theorem. Is this reasoning correct?
Question about contour integration (non-rigorous question).
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complex-analysis
1 Answers
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Yes, that's correct. I don't know why you call this "non-rigorous". It's perfectly rigorous (assuming that the "inner" boundaries consist of curves that you can integrate over: piecewise smooth, say).