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Would it be right to conclude that the closed integral $\int_P z^{1\over 3}dz=0$ for $P$ being the circle $|z-z_0|=|z_0|$ by Cauchy's theorem despite the fact the the integrand is not holomorphic at one point, namely $z=0$, on the path $P$?

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    No, to apply Cauchy's theorem you must have a holomorphic function. Perhaps you can get this result by taking a limit after applying Cauchy's theorem...2012-02-18
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    @GEdgar: Would you mind saying a little bit more on what you mean by taking a limit after applying Cauchy's theorem?2012-02-18
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    You want an integral around a circle passing through $0$. You know an integral around a slightly smaller circle (not inclosing $0$) does have value $0$ by Cauchy's theorem. Try to use some limit argument now, to show the integral on the large circle is the limit of the integral on smaller circles.2012-02-18
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    @GEdgar: Thanks, but doesn't that require some sort of continuity or something of the sort? For example if our bad point is a singularity we can't really take the limit? I might just be confused...2012-02-18
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    Continuity, yes. And a fairly simple application of it. But (in answer to your question) this is not simply an immediate application of Cauchy's theorem.2012-02-19

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