Cauchy's theorem (see 5.7 in Conway's book, p85) tells us the following:
Let $G\subseteq \mathbb{C}$ be an open subset and $f: G \to \mathbb{C}$ an analytic function. If $\gamma$ is a closed rectifiable curve in $G$ such that $n(\gamma,w) = 0$ for all $w\in \mathbb{C}-G$, then $$\oint_\gamma f = 0.$$
Consider now the following paragraph in Conway's book:
How is the sentence $\int_\gamma f = 0$ for all closed curves $\gamma$ true, if $f$ is analytic?
For example, consider $G= \mathbb{C}\setminus \{0\}$. Consider $\gamma(t) = \exp(2\pi it), 0 \le t \le 1$. Then the function $z\mapsto z^{-1}$ is analytic on $G$, $\gamma$ is a closed surve in $G$ and $$\oint_\gamma z^{-1} dz = \int_0^1 \frac{2\pi i \exp(2\pi i t)}{\exp(2\pi it)}dt = 2\pi i\ne 0.$$ What am I missing here?
To re-emphasise the point about $\int_{\gamma}f=0$.
If $f$ is analytic on a domain - an open simply connected set - all integrals of $f$ around a closed contour are zero. Without simple connectivity you allow for singularities to hide in the “holes” of the set, and it may be that the function has nonzero residues hence a nonzero integral around certain closed contours.
$G$ is not a region, in the sense I assume Conway defines (open, simply connected). It does not have simple connectivity due to the “hole” at zero: simple connectivity is important in the proofs of the Cauchy theorems.
The presence, and effect, of the ‘hole’ is touching on the residue theorem. $2\pi i=2\pi i\cdot(1)$ and $(1)$ is the residue of $z\mapsto z^{-1}$ at $0$.