Multiplier of fixed point $\infty$ as limit

Question

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 $\infty$?

Answer 1

$g(z) = 1/(R(1/z))$ 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.