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Why the contour integral of $\,\displaystyle{f(z)=\frac{i-1}{z+i}}\,$ is not zero although it should be because $f(z)$ is analytic? I have used contour $z=\gamma(t)=2e^{it}$, where $0\leq t\leq\pi$.

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    Also: the specified contour is not closed.2012-08-25

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It makes no sense to say a function is analytic on its own - you must also specify an open subset of $\mathbb{C}$ that $f$ is analytic over. Here, $f$ is not analytic inside the specified contour - it has a simple pole at $z=-i$ which is contained inside.

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    @alvoutila : Correct. I.e. with no singularities in the region that it surrounds. If it is holomorphic everywhere in that region and every where in some open set that the contour is in, then the integral is $0$. But this curve fails to be holomorphic at one point in the region that the curve surrounds.2012-08-25