In his "Iteration of Rational Functions" Beardon defines a critical point of a rational function (mapping the Riemann sphere to itself) as a point such that the function is not injective in any neighborhood of that point.
I'm trying to square Beardon's definition with the other definition, in which a critical point is a point at which the function is either not differentiable or the derivative is zero. Specifically, I'm trying to determine when/if/how I can use the second definition to evaluate whether or not $\infty$ is a critical point.
When is $\infty$ a critical point of a rational function mapping the Riemann sphere to itself?
A rule like, "a point $z\in\mathbb{C}$ is critical if $f'(z) = 0$ or $\infty$, and $\infty$ is critical if $1/f'(\infty) = 0$ or $\infty$," would fit nicely with Beardon's classification of fixed (and later periodic) points by multiplier. But I haven't been able to find this rule for critical points stated explicitly in Beardon or anywhere else.
Edit: Thinking about this a little more, I think the rule might be
Infinity is a critical point of $f$ if and only if zero is a critical point of $1/f$.
An argument goes: The valency of $f$ at any point is invariant under conjugation by a Möbius transformation, so instead of considering the valency of $f$ at $\infty$ we can consider the valency of the conjugate $g:z\to 1/f(1/z)$ at $\infty$. Now
$\begin{aligned} \frac{\partial}{\partial z}(1/f(1/z))(\infty) &= \lim_{a\to\infty}{\frac{\partial}{\partial z}}(1/f(1/z))(a)\\ &= \lim_{a\to 0}{\frac{\partial}{\partial z}}(1/f(z))(a)\\ &= \frac{\partial}{\partial z}(1/f(z))(0) \end{aligned}$.
And the second step there is supposed to be justified by the fact that everything in sight is continuous (though I'm afraid I might be forgetting some basic calculus gotcha). Anyway, assuming that the above steps are correct and assuming that knowledge of the derivative is all that's needed to determine the valency, then that should do it: The valency of $f$ at $\infty$ is the same as the valency of $1/f$ at zero.
Just choose an appropriate pair of charts. Write $f(z) = p(z) / q(z)$, where $p$ and $q$ are polynomial functions with no common zeros. We may assume $f$ is not a constant function – the constant case is trivial. There are a few cases:
$f(\infty) = \infty$ – look at the function $w \mapsto \frac{1}{f(1 / w)}$. The derivative is given by $$\frac{f' (1/w)}{w^2 f(1/w)^2} = \frac{p'(1/w) q(1/w) - p(1/w) q'(1/w)}{w^2 p(1/w)^2}$$ but it's not immediate how to extract anything useful from this expression. Instead, suppose the leading term of $p(z)$ is $a z^n$ while the leading term of $q(z)$ is $b z^m$. We must have $n > m$ if $f(\infty) = \infty$, and the numerator is of degree $n + m - 1$ (in terms of $1/w$) and the denominator is of degree $2 n - 2$. Now there are two subcases:
$f(\infty) \ne \infty$ – look at the function $w \mapsto f(1 / w)$. The derivative is given by $$-\frac{f' (1/w)}{w^2} = -\frac{p'(1/w) q(1/w) - p(1/w) q'(1/w)}{w^2 q(1/w)^2}$$ but again it's not clear what we can say from this. Let $a z^n$ be the leading term of $p(z)$ and let $b z^m$ be the leading term of $q(z)$. We must have $n \le m$ if $f(\infty) \ne \infty$, and the numerator is of degree $n + m - 1$ (if $n \ne m$) or $< n + m - 1$ (if $n = m$) and the denominator is of degree $2 m - 2$. Now there are four subcases:
At any rate, the point is that there is no easy criterion purely in terms of the values of $f$ and $f'$.