A small remark after my complex analysis qualifying exam

Time really goes by, I should write more here. I have graduated with a master of science in mathematics at University of Hamburg, enjoyed my last days there with friends and concerts (both as a performer and also listener), and was on a family vacation in China before moving to the US. To make it short, despite the stress, I am enjoying life right now. The people are really nice here, my apartment is functional and I also wrote my analysis qualifying exams today.

Both the complex and the real part went well and were very manageable. One question that came quite often in complex analysis was the following: Let $f$ be a meromorphic function in $\mathbb{C}$ that satisfies $|f(z)|\to\infty$ as $|z| \to \infty$. Show that $f$ is rational.

It should be fairly straightforward:

First, show that $f$ has an isolated pole in infinity. Then, deduce that $f$ has only finitely many poles. Remove all poles $z_0$ in $\mathbb{C}$ by multiplying $f$ by $(z - z_0)^n$ where $n$ is the order of $z_0$. This new function $g$ is entire with a pole at infinity. Look at $g(1/z)$ and conclude with the Laurent series at $z = 0$ that $g$ must be a polynomial.

The problem is though that the first part is not that easy, at least to me. Because I didn’t see it anywhere on the internet, I thought I may write it here for future reference: Since $|f(z)| \to \infty$ for $|z| \to \infty$, there exists $R > 0$ such that $|f(z)| \geq 1$ for all $z \geq R$. Consider now $g(z) = 1/f(z)$ in $\Omega = \{z \in \mathbb{C} \colon |z| \geq R\}$. Clearly, since $f$ is non-zero on $\Omega$, we have $g(z)$ is analytic in $\Omega$ with every pole of $f$ in $\Omega$ corresponding to a zero of $g$. In particular, if $f$ had infinitely many poles in $\Omega$, then $h(z) = g(1/z)$ has infinitely many zeros in $\Omega’ = \{z \in \mathbb{C} \colon |z| \leq 1/R\}$. This would imply that $\{z \colon h(z) = 0\}$ has limit point, meaning $h \equiv 0$ by the uniqueness theorem. This would mean that ‘‘$f(z) = \infty$’’, which obviously contradicts $f$ being meromorphic. So, $f$ has only finitely many zeros in $\Omega$. In particular, we can choose $R’ \geq R$ sufficiently large that $f$ has no poles in $\{z \in \mathbb{C} \colon |z| \geq R’\}$. Hence, $\infty$ is an isolated singularity of $f$.

Funnily enough, I came up with it the night before the exam. I did see the question quite often in older qualifying exams, but didn’t think much of it till it was explicitly given as a hint in one of the newer formulations of the problem. Oh well, hopefully it’s correct.