Show that there is an analytic $g: U \rightarrow \mathbb{C}$ such that $f = e^g$ when $f$ is nonzero and analytic.

Question

Let $U$ be a simply connected region, $U \subset \mathbb{C}$. Let $f: U \rightarrow \mathbb{C}$ be analytic and never $0$. Show that there is an analytic $g: U \rightarrow \mathbb{C}$ such that $f = e^g$.

Answer 1

Let $g$ be a primitive of $\frac{f'}f$. Then$$(fe^{-g})'=f'e^{-g}-f\frac{f'}fe^{-g}=0.$$So, $f=Ke^g$ for some non-zero constant $K$. Choose $k$ such that $K=e^k$ and then $f=e^{g+k}$.