Multiplier of fixed point $\infty$ as limit For $R: \mathbb{C}_\infty \rightarrow \mathbb{C}_\infty$ a rational map with $R(\infty) = \infty$ I know that the Multiplier of $\infty$ under R is:
$$
m(R,\infty) = (\phi \ \circ \ R \ \circ \ \phi^{-1})'(0)
$$
with $\phi(z) \mapsto 1/z$ (Thanks Martin R). Is it true that
$$
m(R,\infty) = \frac{1}{R'(\infty)} = \lim_{z \rightarrow \infty} \frac{1}{R'(z)} \ \ \ ?
$$
I tried to prove or refute it, but I can not handle the $" 0 \ \cdot \infty"$ if I apply the definition. Also does it hold for all holomorphic functions defined at $\infty$?
 A: $g(z) = 1/(R(1/z))$ has the derivative
$$
 g'(z) = \frac{R'(1/z)}{z^2 R^2(1/z)}
$$
so that
$$
m(r, \infty) = g'(0) = \lim_{z \to 0} \frac{R'(1/z)}{z^2 R^2(1/z)}
=   \lim_{z \to \infty} \frac{z^2 R'(z)}{ R^2(z)}
= \lim_{z \to \infty} \left( \frac{z R'(z)}{R(z)}\right)^2 \frac{1}{R'(z)} \, .
$$
Now consider two cases: If $R(\infty) = \infty$ with multiplicity one then
$$
 R(z) = cz + O(1)
$$
for $z \to \infty$ with some complex constant $c \ne 0$. Then $\lim_{z \to \infty} \frac{z R'(z)}{R(z)} = 1$ and therefore
$$
m(r, \infty) = \lim_{z \to \infty} \frac{1}{R'(z)}  \, .
$$
And if $R(\infty) = \infty$ with multiplicity $k \ge 2$ then
$$
 m(r, \infty) = 0 = \lim_{z \to \infty} \frac{1}{R'(z)}
$$
holds as well.
$R$ need not be rational for this calculation, it suffices that $R$ is holomorphic in a domain $\{ z : r < |z| < \infty \} $ for some $r > 0$ and has a pole at $z = \infty$.

Alternatively one can calculate that if $R(\infty) = \infty)$ with multiplicity $k \ge 1$ then
$$
 R(z) = c z^k + O(z^{k-1})
$$
for $z \to \infty$ and
$$
\frac{1}{R(1/z)} = \frac{z^k}{c} + O(z^{k-1})
$$
for $z \to 0$, which also gives the desired result.
