Let $U\subset\Bbb C$, and let $f:U\to\Bbb C$ be holomorphic on $U$ and continuous on $\overline{U}$, with $U$ a simply connected open subset. Let $\gamma$ be a closed curve in $U$, of winding number $1$ about the point $z\in U$. Then:
$$f(z)=\frac{1}{2\pi i}\oint_{\gamma}\frac{f(w)}{w-z}\,\mathrm{d}w$$
Let $A$ be a square matrix, and $I$ the identity matrix of the same dimensions. Then if $\gamma$ encloses all eigenvalues (from Wikipedia and other sources) of $A$, $f(A)$ can be defined as:
$$f(A)=\frac{1}{2\pi i}\oint_\gamma f(w)(wI-A)^{-1}\,\mathrm{d}w$$
And the logic for this is, I believe, that if $w\in\gamma$ were ever an eigenvalue of $A$, the expression $(wI-A)^{-1}$ would not exist by definition and the integral formula would fail to be well defined. However, the sources I read never explained this requirement, so my guess could be way off the mark.
Letting $\gamma$ enclose all eigenvalues so that $w$ is comfortably never an eigenvalue, $\forall w\in\gamma$, is a straightforward way to keep the integral well defined. However, I wonder in the interests of rigour, could I have a path $\gamma$ such that $\forall w\in\gamma,\exists(wI-A)^{-1}$ but some eigenvalues of $A$ are not enclosed. For example, if $A$ had eigenvalues $(1+2i,3,-i)$ and if $\gamma$ was the open disk of radius $\frac{1}{2}$ about the origin, $\gamma$ would enclose no eigenvalues, but no eigenvalues would lie on $\gamma$, and to my mind this would keep the integral well defined, since $A$ isn't exactly a point in the interior of $\gamma$ so $\gamma$'s placement is less important.
Are there any problems here I'm overlooking?
EDIT: One user has pointed out in the comments that there must be at least one singularity in the interior of $\gamma$ for the integral to be non-zero, so my new question is: if we let $\gamma$ enclose only one eigenvalue, so that the integral is non-zero but many eigenvalues are not enclosed, would this still be well-defined or a good generalisation?