behavior of a meromorphic function around an essential singularity

Question

Suppose $f(z)$ is a meromorphic function in a punctured disc $\mathbb D^\times$ and $0$ is an essential singularity of $f(z)$. Is it possible that $f(z)$ takes all values in the extended complex plane $\hat{\mathbb C}$ in any neighborhood of $0$? That is, is it possible that $f(z)$ has a pole and also takes every finite complex value in arbitrarily small neighborhood of $0$?

Answer 1

If we use the definition of "essential singularity" that doesn't require that to be an isolated singularity, then it is possible (and in fact not unusual) that a meromorphic function attains every value in $\widehat{\mathbb{C}}$ infinitely often in every punctured neighbourhood of the essential singularity.

Let's assume the point in question is $0$. By the mapping properties of the exponential function it easily follows that $g(z) = \exp(1/z)$ attains every value of $\mathbb{C}\setminus \{0\}$ infinitely often in every punctured neighbourhood of $0$. Then composing $g$ with a surjective holomorphic $h \colon \mathbb{C}\setminus \{0\} \to \widehat{\mathbb{C}}$ achieves the desired goal (using $h \circ g$). Simple examples of such $h$ are many rational functions (of order $> 1$) like $z \mapsto \frac{z(z-1)}{(z-2)(z-3)}$. So $$f(z) = \frac{e^{1/z}(e^{1/z}-1)}{(e^{1/z}-2)(e^{1/z}-3)}$$

is a concrete example for the behaviour you ask about.

Answer 2

By definition, an essential singularity is an isolated singularity, so if $0$ is one of those there are no poles in some neighbourhood of $0$. There may be infinitely many poles in $\mathbb D^\times$, but they will accumulate at the outer boundary, not at $0$.