Meromorphic function on punctured unit disk with poles at $\frac{1}{n}$ is dense in $\mathbb C$ arbitrarily close to zero

Question

Let $f$ be meromorphic in the punctured unit disk with poles at the points $1/n$, $n =1, 2, 3, \ldots$. I want to show that the image of an arbitrarily small neighborhood of zero under $f$ is dense in $\Bbb C$.

I have reasoned so far:

1. We know (using basic result from infinite products) that we can find $g(z)$ that is has zeros of prescribed orders at exactly $\frac{1}{n}$ and thus we can multiply $h(z) = f(z)g(z)$ and reason that it has an isolated singularity at zero.
2. This isolated singularity has to be essential: if it is removable or a pole, by considering $1/h(z)$ we get contradictions. Specifically, we know that if the singularity gives an expression for $h(z)$ of the form $H(z)/z^m$ s.t. $H(0) \not = 0$. $H(z)$ can only have finitely many zero in any small open disk around $0$. Therefore, considering $1/h(z)$, we get either a sequence of zeros that converges to a pole, a nonzero value or a zero. First two give contradictions by continuity and the last gives a contradiction since it follows that $1/h(z) \equiv 0$.

Now, I am stuck - I don't see how knowing $h(z)$ has an essential singularity at zero allows to conclude $h(z)/g(z)$ has a dense image in $\mathbb C$ near $0$. Any ideas?

I tagged Riemannian geometry but this should be doable using basic complex analysis results.

EDIT: Alternative approach, if $w$ is a value in a neighborhood of which $f(z)$ attains no value, then $1/(f(z)-w)$ has zeros that converge to zero and is bounded near zero, so it must be constant. Does this make sense?

Answer 1

Your second approach works. One can start similarly as in the proof of the Casorati-Weierstrass theorem (which can not be applied directly because $z=0$ is not an isolated singularity of $f$).

Let us denote $$ D_r = \{ z \in \Bbb C \mid 0 < |z| < r \} $$ and $$ G = \{ 1/n \mid n = 1, 2, 3, \ldots \} \, . $$ If the desired conclusion is not true then there are $w \in \Bbb C$, $\epsilon > 0$, $r > 0$ such that $$ |f(z) - w| > \epsilon \text{ for } z \in D_r \setminus G \, . $$ Now define the function $g: D_r \to \Bbb C$ as $$ g(z) = \begin{cases} 1/(f(z) -w) & \text{ if } z \in D_r \setminus G \, ,\\ 0 & \text{ if } z \in D_r \cap G \, . \end{cases} $$ $g$ is holomorphic and bounded in $D_r$, and therefore can be extended to a holomorphic function $\tilde g$ on $B_r(0)$. The identity theorem then shows that $\tilde g$ is identically zero. That is a contradiction because $g(z) \ne 0$ for all $z \in D_r \setminus G$.