Let $f(z) = \frac{p(z)}{q(z)}$ be a rational function. I am interested in determining when $\infty$ is a removable singularity or a pole, and why $\infty$ cannot be an essential singularity of $f$.

I have seen the answer to this question: Elegant proof that a rational function has no essential singularity., and while I understand how the limits were computed in the answer, I am confused how to make the final conclusion.

If $\deg p < \deg q$, then since $\lim_{z\to\infty} f(z) = 0$, I believe that $\infty$ would be removable. This is because that limit implies that there exists $M>0$ such that in the region $|z|> M$, we have that $|f(z)|<N$. That is, $f$ is bounded in some disk centered at $\infty$.

If $\deg p > \deg q$, then since $\lim_{z\to\infty} |f(z)| = \infty$ and so $\infty$ is a pole of $f$.

My issue is with the case where $\deg p = \deg q$. I know that $\lim_{z\to\infty} f(z) = \frac{a_m}{b_m}$, where $a_m$ is the leading coefficient of $p$ and $b_m$ is the leading coefficient of $q$. But why does this mean that $\infty$ is a removable singularity? I don't think that limit means that $f$ is bounded in some punctured disk with center $z_0$. I also tried to show that $f(\frac{1}{z})$ has a removable singularity at $0$, but I have a similar issue.

I realize this must be a basic thing, but I've tried for a long time and cannot figure it out.

  • $\begingroup$ Did you look at some examples, such as, I suppose, $\frac{x^2+1}{x^2+2x+2}$? Your trick of looking at $f(1/x)$ should work, certainly does for that example. $\endgroup$
    – Lubin
    Mar 31 '19 at 2:00
  • $\begingroup$ @Lubin In that case $f(1/x) = \frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+\frac{2}{x}+2}$. Then $\lim_{x\to 0} f(1/x) = 1$, right? Does this mean $0$ is a removable singularity of $f(1/x)$? Why? $\endgroup$ Mar 31 '19 at 2:08
  • 1
    $\begingroup$ Well, that expression simplifies to $g=\frac{1+x^2}{1+2x+2x^2}$, but for $x\ne0$. The fact that its limit at $0$ is $1$ means precisely that when you extend the domain to $0$ so that $g_{\text{new}}(0)=1$, the new function is continuous at $0$. That also is the definition of removable singularity. $\endgroup$
    – Lubin
    Mar 31 '19 at 2:18

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