This answer shows that the Cauchy integral formula yields the analytic power series representation of a matrix function, by uniform convergence of a matrix geometric progression:
$$\sum_{n=0}^\infty\frac{A^n}{z^n}$$
Provided that $|z|\gt|\lambda|,\,\forall\lambda\in\Lambda(A)$ where $\Lambda$ denotes the spectrum of the matrix. The only way I can see for a smooth closed curve $\gamma$ to enclose all eigenvalues and still satisfy $|z|\gt|\lambda|$ is if $\gamma$ is enclose the open disk about the origin $B_r(0)$, where $r=\max_{\lambda\in\Lambda(A)}|\lambda|$, as it is only then that for all $z\in\gamma$ we have $|z|\gt|\lambda|$ for all $\lambda$. However, in reading texts on the Cauchy integral formula for matrix functions, it is always stated that $\gamma$ must enclose $\Lambda(A)$, but never stated that $\gamma$ must have the property I just described.
How can it be that $\gamma$ can just enclose $\Lambda(A)$ without any further requirements? Suppose all the eigenvalues are in one quadrant of the plane - any tightly enclosing $\gamma$ would fail to have the property $\forall z\in\gamma,\,\forall\lambda\in\Lambda(A),\,|z|\gt|\lambda|$ but this property is crucial for the matrix integral formula to be justified as a reasonable extension of a function to a matrix function (due to the uniform convergence property it gives).