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?
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.