for $f$ meromophic on $\Bbb C$ :$f$ is rational <=> $f$ has at worst a pole at ∞

Question

I'm really stuck with this one since ∞ is a very new introduced concept in our complexe analysis

Help is very appreciated, thanks

Answer 1

I think the problem you are encounter with is to see what is a function (holomorphic or meromorphic) looks like at $\infty$. Here is the way, given a function $f$, consider the function $$g(z)=f(\frac{1}{z})$$ and see how $g(z)$ is at $0$. For example $f(z)=z^2$ then $g(z)=\frac{1}{z^2}$. Then $f$ has a pole of order $2$ at $\infty$ because $g$ has pole of order $2$ at $0$. And $f(z)=e^z$ has essential singularity at $\infty$ because $g(z)=e^{\frac{1}{z}}$ has essential singulartiy at $0$.

So with these, I hope you can see by yourself that a rational map

$$f(z)=\frac{a_0+a_1 z+ \ldots + a_n z^n}{b_0 + b_1 z +\ldots+ b_m z^m}$$ has at worst a pole at $\infty$.

Conversely, if a meromorphic function has at worst a pole at $\infty$ then it induce a holomorphic map from $\hat{\mathbb{C}}$ to $\hat{\mathbb{C}}$, hence rational map. The idea is the number of pole is finite $p_1,..p_k$ (possibly $\infty$), with the principle part is $h_1,...,h_k$. Then $$g(z)=f-h_1-\ldots-h_k$$ is holomorphic on $\hat{\mathbb{C}}$ hence constant by maximum principle, which means $$f=h_1+\ldots+h_k+c$$ is a rational map.