Meromorphic Function on Extended Plane

Because $\infty$ is a removable singularity or pole of a meromorphic function $f(z)$, $\exists$ $R>0$, s.t. $f\in H(\{z:R<|z|<\infty\})$.

We note that $f$ has at most finite poles in $\{z:|z|\le R\}$. If not, then the finite poles $\{z_j\}$ must have a convergent sub sequence. Let $z_j\to a(j\to \infty)$. Then $|a|<R$ and $a$ is not a isolated pole, which is impossible.

Let $z_1,z_2,\cdots,z_n$ be all poles of $f$ in $\{z:|z|\le R\}$, with orders of $m_1,m_2,\cdots,m_n$. $f$ has Laurent expansion besides $z_j$, which main part is \begin{equation} h_j(z)=\sum_{k=1}^{m_j}\frac{c_{-k}^{(j)}}{(z-z_j)^k}. \end{equation}

Let $g$ be the main part of $f$'s Laurent expansion besides $\infty$.

If $\infty$ is a pole of $f$, $g$ is a polynomial; if $\infty$ is a removable singularity of $f$, $g\equiv 0$. Note \begin{equation} F(z)=f(z)-\sum_{j=1}^n h_j(z)-g(z), \end{equation} We know that $F\in H(\bf C$ $_\infty$ \ $\{z_1,\cdots,z_n,\infty\}$. The main parts of $f$'s Laurent expansion besides $z_1,\cdots,z_n,\infty$ have been eliminated, thus $f$ is holomorphic. That implies $F\in H(\bf C$ $_\infty$). From Liouville's theorem, we know $F\equiv c$. Then \begin{equation} f(z)=c+g(z)+\sum_{j=1}^n h_j(z) \end{equation} is a rational function.

Let $f$ be meromorphic on $S^2$. Let $A\subset S^2$ be the set of points where $f$ has poles.

Let suppose that $A$ is infinite. Then, by compactness of $S^2$, $A$ must have a limit point in $S^2$ which contradicts the definition of meromorphic functions. Thus $A$ is finite. Let $A=\{p_k\}_{k=1}^n$ and let $d_k$ be the degree of the pole $p_k$ for all $k\in\{1,2,\cdots,n\}$.

Then there exists $g\in H(\mathbb{C})$ s.t. $$g(z)=f(z)-\sum_{k=1}^n\left({\sum_{i=1}^{d_k}}\frac{c_i^{(k)}}{(z-p_k)^i}\right) \text{i.e.} f(z)=g(z)+\sum_{k=1}^n\left({\sum_{i=1}^{d_k}}\frac{c_i^{(k)}}{(z-p_k)^i}\right)\cdots(1)$$

Next, we define $$h(z)=f(z)\cdot\prod_{k=1}^k{(z-p_k)^{d_k}}.$$ Then, using $(1)$ we see that $h$ does not have poles in $\mathbb{C}$ and it may have a pole only at $\infty.$ This means that $h$ is a polynomial, hence, by $(1),$ $f(z)=\frac{h(z)}{\prod_{k=1}^k{(z-p_k)^{d_k}}}$ is a rational function.