In a solution to a problem, I read that, if $f(z)$ is entire, $f(z)\neq0$ and the domain of definition of $f(z)$ is simply connected, then it is possible to choose a branch of log $f(z)$ that is analytic in the entire plane.
I was a bit surprised by this. My understanding is that, given at point $z_0\neq0$, you can always choose a branch so that log $z$ is analytic at that point. But once you have chosen a branch, that branch will not be analytic in the branch cut. To me this implies, that you can never choose $one$ branch of the logarithm, such that log $f(z)$ is analytic in the whole plane.
Grateful if you can sort this out for me. Where do I go wrong? Am I misinterpreting what is said in the solution?
If $f$ is holomorphic on a simply connected domain $U$ and $0\not\in f(U)$ then there exists another function $g$ also holomorphic on $U$ such that $f(z)=\exp g(z)$. In fact, observe that
$$f(z)=e^{g(z)}\implies f'(z)=g'(z)e^{g(z)}=g'(z)f(z)\implies g'(z)=\frac{f'(z)}{f(z)}$$
which inspires $g(z):=\displaystyle \int_w^z\frac{f'(\xi)}{f(\xi)}d\xi$ (this is well-defined since $U$ is simply connected).
It is not however true that $\log f(z)$ can be defined for a continuous single-valued branch of $\log$, for instance consider $f(z)=e^z$ and $U=\Bbb C$ - then $f(U)=\Bbb C\setminus\{0\}$ is not simply connected.