# meromorphic at $\infty$ is a polynomial?

I do not understand the statement why $p$ will be polynomial in the following statement:

"The function $p:\hat{\mathbb C}\rightarrow\hat{\mathbb C}$ defined by $p(z)=f(z)q(z)$ has the removable singularities at the poles of $f$ in $\mathbb C$, so it is entire, this it has power series representation on all $\mathbb C$, also $p$ is meromorphic at $\infty$ as both $f$ and $q$ are.this forces $p$ to be a polynomial.Since, $f=\frac pq$" $$q(z)=\prod_{j=1}^{n}(z-z_j)^{e_j}$$

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What are $f(Z)$ and $q(z)$? –  Makoto Kato Aug 23 '12 at 22:54
You really should answer Makoto Kato's question, in order to make your question comprehensible. But also, notice that \Pi_{j=1}^n gives you $\displaystyle\Pi_{j=1}^n$, whereas \prod_{j=1}^n gives you $\displaystyle\prod_{j=1}^n$. The subscripts are positioned differently when they're in display style, and I think there are also differences when they're in inline style. The latter form is standard. Similar comments apply to \Sigma versus \sum. –  Michael Hardy Aug 23 '12 at 23:37
$F(z)$ is a meromorphic function in extended plane, which has poles at $z_1,\dots z_n$ with multiplicity $e_1,\dots, e_n$ –  miosaki Aug 24 '12 at 20:15

What does it mean to have a pole at $\infty$? This means that the function has a pole when written in terms of the coordinate $w=\frac1z$ .
Or that the function approaches $\infty$ as the argument approaches $\infty$. –  Michael Hardy Aug 23 '12 at 23:33