Show that if $f: \mathbb{C} \rightarrow \mathbb{C} \space \cup \space \infty$ is a meromorphic function in the plane, such that there exists $R, C > 0$ so that for $|z| > R, |f(z)| \leq C |z|^n$, then $f$ is a rational function.


Since $f$ is meromorphic, and $|f(z)| < \infty$ for $|z| > R$, $f$ must have only finitely many poles $a_1, \dots, a_m$ (with multiplicity) in the disk $|z| \leq R$. Let $g(z) = (z-a_1)\dots(z-a_m)f(z)$, then $g(z)$ is entire and $|g(z)| \leq C'|z|^{m+n}$ for $|z|$ large enough, and therefore $g(z)$ is a polynomial of degree at most $m+n$, using Cauchy's estimate $|f^N(0)| \leq C'r^{m+n}N!r^{-N} \rightarrow 0$ as $r\rightarrow \infty$ if $N>m+n$. Thus, $f(z) = g(z)/((z-a_1)\dots (z-a_n))$ must be a rational function.

Question. I'm confused by the application of Cauchy's Estimate.

$$|f^N(0)| \leq |\frac{N!}{2\pi}\oint_C \frac{f(z)}{z^{n+1}} dz| \leq N!r^{-N}|f(z)| \leq \space ???$$

Where the first inequality is by definition of Cauchy Integral Formula. How do you apply the bound on $|g(z)|$ to apparently bound $|f(z)|$ here?

  • $\begingroup$ Could you explain again, why you apply the Cauchy integral formula? Doesnt it suffice to show that g is a polynomial? $\endgroup$
    – klirk
    Aug 9, 2017 at 21:02

1 Answer 1


Cauchy's estimate should be applied to $g$, instead of $f$. You know that, for some constant $C'$, $g(z)\leqslant C'|z|^{m+n}$, if $|z|\gg0$. But then$$\left|\frac{g^{(N)}(0)}{N!}\right|\leqslant\frac{1}{r^N}r^{m+n},$$if $r$ is large enough. Therefore, making $r\to+\infty$, $g^{(N)}(0)=0$ when $N>m+n$. It follows that$$g(z)=\sum_{k=0}^{m+n}\frac{g^{(k)}(0)}{k!}z^k,$$which is a polynomial function.

  • $\begingroup$ Quite Obvious now that you point it out.....appreciate your answers as always on this site Jose. Thanks! $\endgroup$
    – tastykakes
    Aug 9, 2017 at 22:16

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