# If $F$ is entire with removable singularity at $\infty$, then $F$ is constant?

On page 24 of Krantz's Complex Analysis, there is the following proof:

Proposition 2: If $F$ is entire and $F$ has a removable singularity at $\infty$, then $F$ is constant.

Proof: By examining $F(1/z)$, we see that $F$ must have a finite limit at $\infty$. Thus $F$ is bounded. By Liouville's theorem, $F$ is constant.

It's mysterious to me what is meant by the first sentence. What does he mean by "examining $F(1/z)$"? I tried expanding $F$ has a Laurent series around $0$, and then using the fact that $F(1/z)$ has the origin as a removable singularity, but I didn't get the conclusion.

Is the idea to do this?

Let $$F(z)=\sum_{n=-\infty}^{-1}a_nz^n+\sum_{n=0}^\infty a_nz^n$$ be the Laurent expansion around the origin. Then $$F(1/z)=\sum_{n=-\infty}^{-1}a_nz^{-n}+\sum_{n=0}^\infty a_nz^{-n}$$ but since $F(1/z)$ has a removable singularity at the origin, we really have $$F(1/z)=\sum_{n=-\infty}^{-1}a_nz^{-n}+a_0.$$ So $a_n=0$ for $n>0$, and thus $F(z)=\sum_{n=-\infty}^{-1}a_nz^n+a_0$, so $\lim_{z\to\infty}F(z)=a_0<\infty$?

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Note you are essentially considering the transition map between coordinate patches on the Riemann sphere. – Potato Jun 20 '12 at 10:02

Say we have the Taylor expansion around zero: $F(w)=\sum_{n=0}^{\infty} a_nw^n$, then if $w=1/z$ for $z\neq 0$, and by uniqueness of the Laurent expansion we have that this is exactly the Laurent series. Now, a removable singularity at infinity of $F(z)$ means a removable singularity for $F(1/z)$ at $z=0$, but this can only happen if $a_n=0$ for $n>0$.
Your idea is definitely going in the right direction. You've found that $F(z)$ has a finite limit as $|z| \to \infty$, so you can find constants $M$ and $r$ such that $|F(z)| \leq M$ for all $|z| > r$. Now, since $F$ is entire, what can you say about $|F(z)|$ for $|z| \leq r$? Can you conclude that $|F(z)|$ is bounded for all $z$?
Does that just follow from the fact that $|z|\leq r$ is compact, and so $|F(z)|$ achieves a maximum on the disk? Since it's also bounded outside the disk, it must be bounded on all of $\mathbb{C}$. – Hana Bailey May 6 '12 at 5:19