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On the bottom of page 19 of Milne's Modular Forms notes (http://www.jmilne.org/math/CourseNotes/MF.pdf), he says

Let $g$ be a meromorphic function on $S$. After possibly replacing $g(z)$ by $g(z-c)$ we can assume $g$ has neither a zero nor a pole at $\infty$.

Now, in example 1.11 we saw that meromorphic functions on the Riemann sphere $S$ we just meromorphic functions $f(z)$ on $\mathbb{C}$ such that $f(z^{-1})$ is also meromorphic on $\mathbb{C}$. So when he writes $g(z)$ I assume he's talking about $g$ on the chart $S\setminus \{\infty\}$. But the translation $z \mapsto z-c$ doesn't move $\infty$?

Can't we just precompose with some Mobius transform $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ taking $\infty$ to some $a/c$ at which $g$ has neither a zero nor a pole, since such a map is a rational map, as required?

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    You're right. Maybe $g((z-c)^{-1})$ was intended. And yes, precomposing with such a Möbius transformation (or postcomposing with one that maps $\infty$ into $\mathbb{C}^\ast$) is exactly what one does to get rid of a pole in $\infty$. – Daniel Fischer Mar 06 '14 at 15:58

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