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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... – GEdgar Feb 18 '12 at 22:17
@GEdgar: Would you mind saying a little bit more on what you mean by taking a limit after applying Cauchy's theorem? – Frank Feb 18 '12 at 22:20
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. – GEdgar Feb 18 '12 at 22:24
@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... – Frank Feb 18 '12 at 22:32
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. – GEdgar Feb 19 '12 at 3:26

As GEdgar explained in comments, this is not as simple as saying "by Cauchy's theorem". There are several points to be made:

  1. $z^{1/3}$ has a single-valued branch in the slit plane $\{z\in \mathbb C: z/z_0\notin (-\infty,0]\}$.
  2. This branch has continuous (no longer holomorphic) extension to $0$
  3. For $0<r<1$, the integral of $z^{1/3}$ over the circle $|z−z_0|=r|z_0| $ is zero by Cauchy's theorem.
  4. Uniform continuity of $z^{1/3}$ on the set $\{z:|z-z_0|\le |z_0|\}$ allows us to pass to the limit $r\to 1$ under the integral.
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