So we know by Green's Theorem that for $F: \mathbb{R}^2 \to \mathbb{R}^2$, $F = (P,Q)$, with $F$ defined on some open set $U$ such that $\partial U$ is piecewise smooth and simple (see JCT) that we have the relation:
$$ \iint_U \partial_xQ - \partial_yP \mathrm{d}x\mathrm{d}y = \int_{\partial U} F \cdot T \mathrm{d}S$$
This formula is simply the two dimensional stokes theorem. When $F$ is treated as a map from $\mathbb{C} \to \mathbb{C}$, and $P,Q$ is replaced with $u,v$ (denoting the real, imaginary components of the function), we have an analogous result for functions of a complex variable. Specifically, we can write a complex line integral as:
$$
\int_\gamma f (z) \mathrm{d}z = \int_{\gamma} [u(z) + iv(z)] (\mathrm{d}x + \mathrm{id}y)
$$
Then, computing the product above allows us to apply Green's theorem, and the cauchy riemann equations (consequence of holomorphicness,we assume $f$ is holomorphic of course) lead to the double integrals being $0$. However, I am uneasy about this proof for a few reasons. First, where does the decompozition of the form $\mathrm{d}z$ come from? I have never studied complex valued forms/forms over the complex numbers, so maybe this the source of confusion for me. Is it simply the definition? I.e is $\mathrm{d}z$ a convention for $\mathrm{d}x + i \mathrm{d}y$?ARe the forms $dx,dy$ real valued or complex valued?
Note that we need to parametrize $ \gamma$ in
$$\int_\gamma f (z) \mathrm{d}z = \int_{\gamma} [u(z) + iv(z)] (\mathrm{d}x + \mathrm{id}y)$$
to get $$\gamma (t) =x(t)+i y(t)$$
The definition of complex integral is $$\int_\gamma f (z) \mathrm{d}z = \int _{t_0}^{t_1} f(\gamma (t))\gamma ' (t) dt$$
Substituting the parametrization in $$\int_{\gamma} [u(z) + iv(z)] (\mathrm{d}x + \mathrm{id}y)$$ will result in $$\int _{t_0}^{t_1} f(\gamma (t))\gamma ' (t) dt$$