Ahlfors' proof of Cauchy's theorem in a disk

I'm stuck in two parts of Ahlfors' proof of Cauchy's theorem in a disk (page 113), that is, if $f$ is holomorphic in an open disc $D$ then $\int_\gamma f(z)dz=0$ over every closed curve $\gamma$ in $D$.

First part:

Fix $z_0\in D$. We define $F(z)=\int_\sigma f(\zeta)d\zeta$ where $\sigma$ is the path joining $z_0$ with $z$ by taking an horizontal line from $z_0$ and getting to $z$ with a vertical line (hope it is clear).

"It is immediately seen that $\frac{\partial F}{\partial y}(z)=if(z)$". Not for me. I mean, I'm geometrically and intuitively inclined to understand that when we derive vertically, since it is a constant path what we should get is the value of $f$. I don't see how the $i$ appears, though, and I formally don't understand what's going on.

First of all, what does $\frac{\partial F}{\partial y}$ mean? This is what I understand by that: if $F=u+iv$ then $\frac{\partial F}{\partial y}=\frac{\partial u}{\partial y} + i \frac{\partial v}{\partial y}$. Right?

I also found the following formula scribbled on my notebook: $\int_\gamma f(z)dz = \int_\gamma f(z) dx + i \int_\gamma f(z) dy$. Is this correct? I don't see how it makes sense to integrate with respect to $x$ a complex-valued function: what am I supposed to do with the $y$'s in the integrand? I'm guessing some abuse of notation is going on here.

Anyway, I'm guessing this is something easy and I'm just confused by notation.

Second part: we get that $\frac{\partial F}{\partial y}(z)=if(z)$ and $\frac{\partial F}{\partial x}(z)=f(z)$. "Hence $F$ is holomorphic with derivative $f$". How is that? I mean, we have $\frac{\partial F}{\partial x}= -i \frac{\partial F}{\partial y}$, which does not seem like Cauchy-Riemann to me. I'm guessing this is the same confusion as above.

Hope I was not overly verbose.

-
$f(z)=u(x)+iv(y)$ is incorrect. Both $u$ and $v$ are functions of both $x$ and $y$. –  Jonas Meyer Jan 27 '11 at 2:28
In particular, $f(z) = u(x,y)+iv(x,y)$. –  PEV Jan 27 '11 at 2:32
Oh yes, sorry. Fixed. –  lentic catachresis Jan 27 '11 at 2:41
In the imaginary direction $dz=idy$ so $\frac{\partial}{\partial z}=\frac{1}{i}\frac{\partial}{\partial y}$, hence $\frac{\partial F}{\partial y}=i \frac{\partial F}{\partial z} = i f(z)$. –  ulvi Jan 27 '11 at 2:42
@ulvi: Again, I'm inclined to believe that, as well that in the real direction, $dz=dx$, but I don't formally understand it. –  lentic catachresis Jan 27 '11 at 18:44

Second part: The Cauchy-Riemann equations are equivalent to $\frac{\partial F}{\partial y}(z) = if(z)$. This form dictates when a function is conformal. In particular, look at the matrix $$\begin{bmatrix} u_x & -v_x \\ v_x & \ \ \ u_x \end{bmatrix}$$