Is the following assertion true?
Let $f:F\rightarrow\mathbb{C}$ be an analytic function, is it true that $\log f(z)$ has an analytic branch if and only if $f$ is zero free and $f(F)$, the image of $F$ under $f$ does not have a hole containing the origin?
I think this will be the case since the the logarithm is analytic in a domain $F$, if and only if $F$ does not contain zero and $F$ does not have a hole containing zero (so that we can always choose the branch cut to stay away from $F$).
I guess I am also using the following claim which seems to be true. Let $f$ and $g$ be analytic functions, then $f(g(z))$ is analytic on a set $F$ if and only if $g(z)$ is analytic on $F$ and $f(z)$ is analytic on $f(F)$.
Are my claims correct? Any help is appreciated, thanks!
I'm not quite sure what exactly you mean by "a hole". But you might consider the case of $f(z) = \exp(z)$ with, say, $F = \mathbb C$, where $f(F)$ is the complex plane with $0$ removed, and yet $\log(f(z))$ has an analytic branch, namely $z$.