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

complex-analysismeromorphic-functionssingularity

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?

Best Answer

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 $z \in D_r \setminus G$.

Related Question