The contour integral around a circle of radius R with function as in the title is given to be $ 2 \pi i $. However the circle is a close contour, why isn't it 0?
Closed contour integral of function $ f(z)=\frac{1}{z} $
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complex-analysis
3 Answers
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Cauchy's integral theorem only applies if $f$ is holomorphic over a simply connected region that contains the contour. For our $f(z)$, this is not the case.
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$f(z) = \frac{1}{z}$ is not analytic at $z = 0$. That's mainly the reason.
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You're circleling a pole (at $z=0$). Therefore it is not possible to contract the contour to a single point, you MUST evaluate the integral. By getting the Laurent-Series of the integrand and getting the coefficient of $z^{-1}$ and multiply it by $2 \pi i$.