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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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    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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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, 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.