Let $f: \mathbb{A} \to \mathbb{C}$ be defined as above, and $C$ be the circle of radius $1/2$ centred at the origin. We want to show that the integral
$$ \oint_C f(z) dz = \oint_C \frac1{z} dz + \oint_C \frac{z}{1-z^2}dz$$
is nonzero. Firstly, the value of the integral only depends on the values $f$ takes along $C$, and so for the purposes of evaluating the integral we can forget that $f$ used to be defined on $\mathbb{A}$, and treat it as being defined on (a neighbourhood of) $C$.
By applying Cauchy's integral formula to the function $g(z) = 1$ with $z_0 = 0$, on the simply-connected domain $\mathbb{C}$, we can find that
$$2 \pi i = \oint_C \frac1{z} dz$$
Since the value of the contour integral only depends on the values that $1/z$ take along the circle $C$, this result is still valid in our case.
For the remaining integral, notice that the function $h: \mathbb{D} \to \mathbb{C}$, $h(z) = \frac{z}{1-z^2}$ (where $\mathbb{D}$ is the open unit disk centred at the origin) takes the same values as the integrand along the curve $C$. Therefore,
$$ \oint_C h(z) dz = \oint_C \frac{z}{1 - z^2} dz$$
Since $h(z)$ is analytic on the simply-connected domain $\mathbb{D}$, by Cauchy's integral theorem the integral is $0$.
It would be instructive to consider the case of a region $D$ with a $2$-dimensional "hole" in it. Taking a point $z_0$ in this hole, we would not be able to construct a curve which "goes to $\infty $" and which stays within $\epsilon $ of $D^c$. Try drawing a picture... This, intuitively, is because $\gamma (t)$ would have to pass through $D$ to get to $\infty$, thereby creating some separation between $\gamma (t)$ and $D^c$, for certain $t$..
Best Answer
Observe that the index of $\;z_0\;$ wrt the closed, simple curve $\;\gamma: |z-z_0|=\frac32\;$ is not zero, since
$$n(\gamma,z_0):=\frac1{2\pi i}\oint_{\gamma}\frac{dz}{z-z_0}=1$$
and from here that $\;A\;$ isn't simply connected.