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

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 $\infty$, and is thus a polynomial, so $f$ is a rational function.

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

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  • $\begingroup$ @HenningMakholm Thanks for that example. $\endgroup$ Commented Mar 22, 2012 at 13:20
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    $\begingroup$ By definition, the only singularities for meromorphic functions are poles, so if the function is indeed meromorphic on the Riemann sphere (and not just on $\mathbb{C}$), the singularity at $\infty$ must be (removable or) a pole. $\endgroup$
    – mrf
    Commented Mar 22, 2012 at 14:30

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

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

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  • $\begingroup$ Thanks again. Do you know what happens if $f$ doesn't even have a singularity at $\infty$? $\endgroup$ Commented Mar 22, 2012 at 13:26
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    $\begingroup$ Then the restriction to $\mathbb C$ has a removable singularity at $\infty$ instead, and so must be rational. $\endgroup$ Commented Mar 22, 2012 at 13:29
  • $\begingroup$ Henning Makholm: Why does the $f$'s restriction to $\mathbb{C}$ imply that $f$ has a removable singularity at $\infty$? $\endgroup$ Commented Apr 28, 2014 at 19:52
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    $\begingroup$ @user1770201: The premise of that comment was that the entire $f$ doesn't have a singularity at $\infty$. In that case $f|_{\mathbb C}$ does have a singularity at $\infty$ (by definition, because it is not defined there), but it is removable -- you remove it by adding the known value of $f(\infty)$ back in! $\endgroup$ Commented Apr 28, 2014 at 21:43

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