Let $f(z)$ be analytic in the whole plane, and suppose that $f(z)$ has a nonessential singularity at $\infty$, Prove that $f(z)$ reduces to a polynomial.

My Thoughts so far :

Since $\infty $ is not an essential singularity of $f$ one of the following can happen

1) $\lim_{z \rightarrow \infty} f(z) = \infty $ ($\infty$ is a pole of finite order)


2) $\lim_{z \rightarrow \infty} f(z) = a \in \mathbb{C} $ ($\infty$ is a removable singularity)

Because $f$ has no poles , 2) implies that $f(z)=c, $ and we are done.

But in case 1) should I try to somehow show that $f^{(n)}(a)$ vanishes for some $n \in \mathbb{N}$ and for all integers $k > n$ ? Cauchy's estimate seems not to be helpful .

On the other hand, we can say that the behavior of $g(z)=f(\frac{1}{z})$ around zero is the same as the behavior of $f(z)$ at $\infty$ and Because $\infty$ is a pole of finite order, Can I say that $g(z)=\frac{h(z)}{z^k}$, where $\lim_{z \rightarrow 0} h(z) \neq 0 \ \ \ $ AND ? $ \ \ \ \lim_{z \rightarrow 0} h(z) \neq \infty $ Hence


According to above , Can we conclude that $f(z)$ is a polynomial ?

Thank you in advance.


If $|f(z)|\to \infty$ as $|z|\to \infty$ then look at the function $g(z)=1/ f(1/z)$ which has a removable singularity at $0$ and $g(0)=0$. If $m$ is the order of the zero of $g$ at $0$ then there exists an entire function $h$ which doesn't vanish around $0$ such that $$g(z)=z^m h(z)\Rightarrow f(1/z)= \frac{1}{z^m} \frac{1}{h(z)}$$ for $z$ in a neighborhood of $0$. Since $h$ is non-zero around $0$, $1/h$ is holomorphic around $0$, so it is bounded in some ball $|z|\leq R$. Hence, $1/|h(z)| \leq C$ for $|z|\leq R$ and by the above $$|f(1/z)|\leq C/ |z|^m$$ for $|z|\leq R$. Equivalently $$|f(z)|\leq C|z|^m$$ for $|z|\geq R'=1/R$. Now using Cauchy estimates for the derivatives of entire functions you can prove that $f$ is a polynomial of degree at most $m$.

  • $\begingroup$ you mean $$g(z)=z^m h(z)\Rightarrow f(1/z)= \frac{1}{z^m} h(\frac{1}{z})$$ right ? $\endgroup$ – the8thone Dec 18 '13 at 3:16
  • 1
    $\begingroup$ No, I just took reciprocals. Look at the definition of $g$. $\endgroup$ – Dimitris Dec 18 '13 at 3:18
  • $\begingroup$ I don't want to edit this answer, but the penultimate inequality holds when $0 < |z| \leq R$, and the last one holds when $|z| \geq 1/R$. $\endgroup$ – user40167 Dec 8 '14 at 4:45
  • $\begingroup$ That's right, but the key idea is that there exists such an $R$. $\endgroup$ – Dimitris Dec 8 '14 at 8:57

If $f$ has a pole of finite order at infinity, take the Laurent series, and subtract the "principal" part, which is a polynomial. Then you are reduced to your case 2.

  • $\begingroup$ In the series $\sum^{\infty}_{k=-\infty} a_k (z-z_0)^k$ , don't we require that $z_0 \in \mathbb{C}$ ? can we allow $z_0$ be in the extended complex plane (Riemann Sphere) ? $\endgroup$ – the8thone Dec 18 '13 at 3:07
  • 1
    $\begingroup$ You have to transform the function to get $z_0$ finite, for example by $z\rightarrow 1/z.$ then subtract the principal part, then transform back. After all, what does "pole at infinity" mean? $\endgroup$ – Igor Rivin Dec 18 '13 at 3:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.