8
$\begingroup$

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

  • 1
    Note you are essentially considering the transition map between coordinate patches on the Riemann sphere.2012-06-20

2 Answers 2