If is analytic on an (non-trivial) open set and for all point then for some particular definition of . Thus, our goal is to prove that as you defined it above has the property that but then that would mean for some particular definition of of the argument, but then this contradicts the fact that is analytic on because the branch of will have to hit the annulus as it leaves the origin. There is just one detail I do not know how to show, maybe it is not needed: there is a point on the annulus so that . With that we can complete the proof. Suppose that and consider then and this means for some complex number because the annulus is a region (open connected set). This means, for all point on the annulus. In particular for , which yields. . Thus, . We have show is the inverse function of and this completes the proof using the begining comments.