# An analytic function reduces to a polynomial

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)

or

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

$f(z)=g(\frac{1}{z})=z^kh(\frac{1}{z})$

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

Thank you in advance.

## 2 Answers

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 $m$.

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

• 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) ? – the8thone Dec 18 '13 at 3:07
• 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? – Igor Rivin Dec 18 '13 at 3:19