Fundamental theorem of calculus for analytic functions I am confused about the proof in my literature of the fundamental theorem of calculus for analytic functions.
The theorem is that if $f(z)$ is continuous on some domain $D$, and $F(z)$ is a primitive for $f(z)$, then $$\int_A^B f(z) dz = F(B) - F(A), $$ over any path in $D$ from $A$ to $B$.
In my literature, they go on to say that this follows from the "regular" fundamental theorem of calculus. Then they say the following:
Since $$F'(z) = \frac{\partial F}{\partial x} + \frac{1}{i}\frac{\partial F}{\partial y},$$ we have that $$F(B)-F(A) = \int_A^B dF = \int_A^B \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy $$
$$= \int_A^B F'(z)(dx+idy) = \int_A^B f(z) dz. $$
I don't really understand what they just showed there. I don't understand if they start by saying that it follows from the "regular" version of the theorem, and that the other part shows something else (perhaps relating to how this holds independently of which path we've chosen) or if they have shown HOW it follows from the "regular" version. And either way, I don't see it. Since I am teaching myself, I am hoping that someone can help fill in the blanks.
(Sorry that this is such a badly formulated question, but it is because that is how confused I am, that I can't even ask a good question about it.) 
 A: The second part is the proof of the theorem.  Here what they refer to as the "regular" Fundamental Theorem of Calculus is a version of the Fundamental Theorem of Calculus for integrals of one-forms on parametrized curves.  Namely, if $\gamma:[a,b]\to D$ is a curve and $F:D\to\mathbb{C}$ is a function, then $$\int_\gamma dF=F(\gamma(b))-F(\gamma(a)).$$ (Here you need $\gamma$ and $F$ to be sufficiently nice, say $C^1$, for this to make sense.)  If you unravel the definition of $\int_\gamma dF$, you find that it is equal to just $\int_a^b g'(t)\,dt$ where $g(t)=F(\gamma(t))$, so this parametrized Fundamental Theorem of Calculus follows from the ordinary Fundamental Theorem of Calculus on $\mathbb{R}$.
In particular, then, if $\gamma(b)=B$ and $\gamma(a)=A$ and we denote the integral along $\gamma$ by $\int_A^B$, we get $$F(B)-F(A)=\int_A^B dF.$$  The rest of the derivation is then just an evaluation of the $1$-form $dF$ to show that it is equal to the $1$-form $f(z)\,dz$, so that $$F(B)-F(A)=\int_A^B dF=\int_A^Bf(z)\,dz.$$ Note that the equation $$F'(z) = \frac{\partial F}{\partial x} + \frac{1}{i}\frac{\partial F}{\partial y}$$ is wrong and should instead be $$F'(z) = \frac{\partial F}{\partial x} \color{red}= \frac{1}{i}\frac{\partial F}{\partial y}$$ which may be the source of some of your confusion.
