If $f$ is a $C^2$ function on an open set $A \subset \mathbb{C}$ containing $z=0$ and we define $$g(z_1):=\int_{0}^{z_1}F(z) dz + f(0),$$ where $z_1 \in A$ and $F(z):= \frac{\partial f}{\partial z}(z)$ is holomorphic, does this imply that
$$dg=\frac{\partial f}{\partial z} dz?$$
I saw this equivalence on a proof but I don’t know how to rigorously deduce it as I haven’t worked yet with total differentials in complex analysis. If that’s of any use, I know that we can prove that $g’(z)=f’(z)$.
Use the decomposition of the total complex derivative $d=\partial + \bar{\partial}$, i.e., $$dh(z) = \frac{\partial h}{\partial z}dz + \frac{\partial h}{\partial \bar{z}}dz $$ for every $h$ differentiable (in the sense that both partial derivative with respect to $x$ and $y$ exists where $z=x+iy$). Then remember that $$\frac{\partial h}{\partial \bar{z}}=0$$ for every holomorphic function $h$. Finally use Stokes' Theorem: $$\int_{\partial \Omega} w = \int_{\Omega}dw$$ with $w=f(z)$ and $\Omega$ the image of the curve (in $A$) that joins $0$ and $z_1$.