Let $D$ a connected open set of $\mathbb{C}$. A continuous function $f:D\to \mathbb{C}$ is a branch of log if $e^{f(t)}=t$ on $D$.
In my book (Cartan) it is written that if $F$ is a primitive of the 1-form $dz/z$ on $D$ then $F$ is a branch of log on $D$.
It is easy to show that if $f$ is a branch of log on $D$, then $f'(t)=1/t$. So I'd like to prove the converse statement. Any suggestions?
The converse is actually false. For example, define $f(z)=\int_1^z\frac{dt}{t}+1$ in the open disc $D=\{z\in\mathbb{C}:|z-1|\lt 1\}$. It is clear that $\int_1^z\frac{dt}{t}$ is a branch of the logarithm on $D$, so $$ e^{f(z)}=e^{\int_1^z\frac{dt}{t}}e^1=ez\neq z. $$
However, the assertion is true with the additional assumption that $e^{f(t)}=t$ for some $t$ (e.g. if $f(1)=0,2\pi i,\dots$). Let $g(t)=te^{-f(t)}$ on $D$, so that $$ g'(t)=e^{-f(t)}-tf'(t)e^{-f(t)}=e^{-f(t)}\left(1-tf'(t)\right)=0. $$ This shows that $g$ is constant on $D$, i.e. $e^{f(t)}=kt$ for some constant $k\in\mathbb{C}$. Assuming $e^{f(t_0)}=t_0$, we have $k=1$ as desired.