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 $0$.
Cauchy's Theorem was earlier, and less refined. Cauchy's Theorem assumed the function was continuously differentiable in a simply-connected region, and it was then proved that all integrals $\oint_{C}f(z)dz$ over simple closed paths $C$ must be $0$. The proof basically relied on Green's Theorem. Goursat's version only required the derivative to exist at each point of the region, without any requirement of continuity of the derivative, and yet the conclusion was the same. Goursat's version allows one to prove that the function has a continuous derivative, without assuming it.
Best Answer
The version I know of this theorem states only the hypothesis that $f$ has a (complex) derivative, except possibly at a finite number of points.
Furthermore, the proof that, if $f$ is holomorphic, it is infinitely differentiable depends on this theorem.