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The Cauchy Integral Formula that I am working with says:

Suppose that $f:E \rightarrow \mathbb{C}$ is holomorphic, $E$ is an open subset of $\mathbb{C}$, and $z_0 \in E$. Pick $\rho > 0$ such that $\overline{\mathbb{B}(z_0, \rho)} \in E$. Let $C \subset E$ be a closed curve such that $\exists$ region $\Omega \subset E \setminus \{z_0\}$ such that $\partial \Omega = C \cup (-\partial \mathbb{B} (z_0, \rho))$. Then $f(z_0) = \frac{1}{2\pi i}\oint \frac{f(z)}{z-z_0}\mathop{\mathrm{d}z}$ (integral is oriented counter clockwise but I can't get the $\LaTeX$ code \ointctrclockwise to work).

We also have the condition on region orientation. In this case, we're taking the convention that the tangent of the region is equal to the normal of the region rotated 90∘ anticlockwise, where the normal points out of the region.

What's stopping the curve $C$ from being contained inside the closed ball?

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  • $\begingroup$ $\rho$ is as small as you may want it to be, so choose a ball inside the region... $\endgroup$ – Alan Muniz Apr 2 '16 at 16:21
  • $\begingroup$ @AlanMuniz That isn't what I'm asking, but I've resolved the problem now. Thanks for the effort anyway! $\endgroup$ – Irregular User Apr 2 '16 at 16:50
  • $\begingroup$ What was the problem? $\endgroup$ – Alan Muniz Apr 2 '16 at 16:52
  • $\begingroup$ @AlanMuniz The problem was that I couldn't see why the case where the curve $C$ is contained inside the closed ball couldn't occur (being able to choose a $\rho$ such that this does not happen is not enough to stop us from choosing one that does make it happen). It turns out that this was to do with curve orientation. $\endgroup$ – Irregular User Apr 2 '16 at 16:57
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Notice that with the convention, we are forced to take the wrong orientation on the curve $C$ and the closed ball.

So a $C \subset \overline{\mathbb{B}(z_0, \rho)}$ that satisfies the hypotheses of the Cauchy Integral Formula is not possible.

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