nidantasnu

Answered

2022-07-09

I know that an analytic function on $\mathbb{C}$ with a nonessential singularity at $\mathrm{\infty}$ is necessarily a polynomial.

Now consider a meromorphic function $f$ on the extended complex plane $\hat{\mathbb{C}}$. I know that $f$ has only finitely many poles, say ${z}_{1},\dots ,{z}_{n}$ in $\mathbb{C}$. Suppose also that $f$ has a nonessential singularity at $\mathrm{\infty}$.

Then if ${z}_{i}$ have orders ${n}_{i}$, it follows that $\prod (z-{z}_{i}{)}^{{n}_{i}}f(z)$ is analytic on $\mathbb{C}$, and has a nonessential singularity at $\mathrm{\infty}$, and is thus a polynomial, so $f$ is a rational function.

But I'm curious, what if f doesn't have a singularity at $\mathrm{\infty}$, or in fact has an essential singularity at $\mathrm{\infty}$ instead? Is $f$ still a rational function?

Now consider a meromorphic function $f$ on the extended complex plane $\hat{\mathbb{C}}$. I know that $f$ has only finitely many poles, say ${z}_{1},\dots ,{z}_{n}$ in $\mathbb{C}$. Suppose also that $f$ has a nonessential singularity at $\mathrm{\infty}$.

Then if ${z}_{i}$ have orders ${n}_{i}$, it follows that $\prod (z-{z}_{i}{)}^{{n}_{i}}f(z)$ is analytic on $\mathbb{C}$, and has a nonessential singularity at $\mathrm{\infty}$, and is thus a polynomial, so $f$ is a rational function.

But I'm curious, what if f doesn't have a singularity at $\mathrm{\infty}$, or in fact has an essential singularity at $\mathrm{\infty}$ instead? Is $f$ still a rational function?

Answer & Explanation

Bruno Dixon

Expert

2022-07-10Added 14 answers

The exponential function has an essential singularity at $\mathrm{\infty}$, but is meromorphic everywhere else -- and is not rational.

On the other hand, if the singularity at $\mathrm{\infty}$ is removable, then the proof you sketch still works, and shows that $f$ must be a rational function.

On the other hand, if the singularity at $\mathrm{\infty}$ is removable, then the proof you sketch still works, and shows that $f$ must be a rational function.

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