Complex analysis primitive of a form: clarification about a proof I have two very very basic question about some proof. 
Theorem: A $C^1$ function $F$ is a primitive for $f(z) ~dz$ if and only if $F'=f$.
Proof: $F$ is a primitive of $f(z) ~dz$ $\Leftrightarrow dF = F_z ~dz + F_{\overline z } ~ d\overline z  = f(z) \Leftrightarrow$ 
$$
\begin{align*}
  F_{\overline z}  &= 0  \\ 
  F_z  &= f(z) 
\end{align*} 
$$
But (here comes the two questions)  $F_z =F'$ so $F'= f(z)$.
The two questions are:
(1) How can the author guarantee that $F$ is holomorphic (has derivative $F'$)?
(2) If $F'$ exists, it is always true that $F_z = F'$. 
Remember that $F_z$ is defined by: $F_z  = \frac12 (F_x  + iF_y)$.
 A: You mess up with $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\overline{z}}$. The correct definition is $\frac{\partial}{\partial z}=\frac{1}{2}(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y})$ and $\frac{\partial}{\partial \overline{z}}=\frac{1}{2}(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y})$. (The way for me to remember is: $\frac{\partial}{\partial z}(z)=\frac{1}{2}(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y})(x+iy)=1$, and $\frac{\partial}{\partial \overline{z}}=\overline{\frac{\partial}{\partial z}}$) Hence, 
$F_z=\frac{1}{2}(F_x-iF_y)$ and $F_{\overline{z}}=\frac{1}{2}(F_x+iF_y)$. (See here)
For question (1), we know that $F$ is holomorphic because $F_{\overline{z}}=0$. To see this, note that $F_{\overline{z}}=\frac{1}{2}(F_x+iF_y)$. So if we write $F=U+iV$, we have $F_x=U_x+iV_x$ and $F_y=U_y+iV_y$, which implies that 
$$F_{\overline{z}}=\frac{1}{2}(F_x+iF_y)=\frac{1}{2}(U_x+iV_x+iU_y-V_y).$$
Therefore, if $F_{\overline{z}}=0$, by the above equality
$$U_x=V_y\mbox{ and }U_y=-V_x,$$
i.e. $F=U+iV$ satifised the Cauchy-Riemann equation. So $F$ is holomorphic. 
For question (2), the answer is yes. If $F'$ exists, i.e. $F$ is holomorphic, then $F'$ should be equal to $F_z$. 
A: I interpret your question as follows: We are given a complex one-form
$$\omega:=a(x,y)dx + b(x,y) dy$$ of the special form
$$\omega=f(z)\ dz\ ,\quad{\rm i.e.,}\quad a=f\ , \ b = i f \qquad(1)$$
for some complex-valued function $f=u+iv$.
Such a form may or may not be the differential of a complex valued function $F$ of the variables $x$ and $y$, called a primitive (or "potential") of $\omega$. The function $F=U+iV$ is a primitive of $\omega$ iff
$$F_x=a=f\ , \quad F_y=b=i f\qquad (2)$$
or
$$U_x+iV_x=u+iv\ ,\qquad U_y+iV_y=-v + i u\ .$$
In particular one necessarily has $U_x=u=V_y$ and $V_x=v=-U_y$. 
It follows that such an $F$ would satisfy the Cauchy-Riemann equations and so would have to be a holomorphic function of the complex variable $z=x+iy$ . In this case $f=F_x=F'(z)$, whence this is possible only if $f$ is holomorphic to begin with. 
Conversely: If we have $(1)$ with a holomorphic $f$ then locally $f=F'$ for a holomorphic function $F$, and $F_x=f$, $F_y=i\, f$ as required by $(2)$.
