Let $A(F)$ be defined as the class of functions that are Analytic in $F^o$ and continuous on $F$.
I am reading a paper with the following assertion:
Let $F$ be an Arakelyan set. Then for any $f\in A(F)$, that is also zero free on $F$, there exists an analytic branch of $\log f(z)\in A(F)$. However, I am thinking of the following counterexample in my mind which must be false but I can't see why.
Let $F=[0,2\pi]$ and $f=e^{i\theta}$, then clearly $f$ is zero free on $F$ and the image of $F$ under $f$ is the unit circle, therefore $\log f(z)$ cannot be in $A(F)$, since any branch will definitely cross the circle causing a discontinuity at that point.
Does anybody know why this counterexample is false? Any help is appreciated, thanks!