Let $f:G\to\mathbb{C}$ denote a holomorphic function over a star-shaped domain $G$ and $f\ne 0$ on $G$. I want to show that it holds
- $\frac{f'}{f}$ is holomorphic
- There is a holomorphic function $h:G\to\mathbb{C}$ such that $e^{h(z)}=f(z)$ for all $z\in G$
(1): Since $f$ is holomorphic, $f'$ is holomorphic too. Moreover $z\mapsto \frac{1}{z}$ is also holomorphic on $f(G)$. So the composition $\frac{f'}{f}$ should be holomorphic as well - no other arguments needed, right?
(2): A hint asks to determine all entire functions $f,g$ with $$f^2+g^2=1$$ and to show $f+ig=e^{ih}$ with an entire function $h$. I think the first equation has something in common with $\sin^2+\cos^2=1$, but I can't put the pieces together and find the relation to the actual question.
EDIT: According to Daniel Fischer's comment: Since $f$ is a holomorphic function on a star-shaped domain $G$ $$F:G\to\mathbb{C}\;,\;\;\;\int_{[z_0,z]}f(\zeta)d\zeta$$ (where $z_0$ is the center of $G$) is a holomorphic function with $F'\equiv f$ on $G$. So, I think we should have $$e^{h}=f\Leftrightarrow\int e^{h(z)}dz=F(z)$$