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