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Does there exist an entire function on $\mathbb{C}$ mapping an open disc to an annulus?

The reason I ask this is because I want to to answer this question: Suppose $f$ is entire, and suppose there exists an open disc $U$ and $\delta > 0$ such that for all $z\in U$, $|f(z)|>\delta > 0$. Does there exist a branch cut for $\log$ such that $\log f(z)$ is analytic on all of $U$? (Note that $f$ is not necessarily one-to-one).

A problem arises if $U$ is mapped to an annulus.

EDIT: I guess the answer to the first question should be no, since if $f$ is entire and nonconstant, it is an open map. Therefore, all interior points of $U$ remain interior for $f(U)$. Thus $\partial f(U) = f(\partial U)$. On the other hand, since $\partial U$ is connected, $f(\partial U)$ cannot be the boundary of an annulus (which consists of two connected components). What do you think?

PS. For a set $S$, by $\partial S$ I of course mean the boundary of $S$.

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up vote 2 down vote accepted

You don't actually need an exact annulus for this, just something where the image of $U$ wraps around a hole and overlaps itself.

Consider, for example, what $z\mapsto e^z$ does to a disc with arbitrary center and radius $>\pi$. Or $z\mapsto z^3$ and a disc of radius $1$ centered at $1+\varepsilon$.

(Your conclusion that $\partial f(U)=f(\partial U)$ is wrong. Just because $f$ maps interior points of $U$ to interior points of its image, it doesn't follow that $f$ must map points outside $U$ to points outside the image of $U$. Therefore $\partial f(U)$ can be a proper subset of $f(\partial U)$).

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