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 $\lim_{z\to\infty}F(z)=a_0<\infty$?
Best Answer
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 $z$?