Q:

Critical points of a function

The literature on Mandelbrot and Julia sets mentions the phase "critical point" quite a lot, but usually doesn't bother to define what it means.
As best as I can tell, a critical point is just any point where the function's derivative is zero.
This cannot be the whole story, however. For example, the same literature often states that the critical points of $f(z) = z^2 + c$ are "zero and infinity".
Clearly $f'(z) = 2z$. Obviously $f'(0)=0$. So zero is a critical point. But how on Earth do they figure that infinity is a critical point? $f'(\infty) = 2\infty = \infty \not= 0$. (??)

Edit: According to Wikipedia, a critical point is anywhere the function's derivative is zero or undefined.
Is this going to be one of those things involving the complex plane "with a point at infinity"? I'm thinking, perhaps as $z$ approaches $+\infty$ then $2z$ does to $+\infty$, but as $z$ approaches $-\infty$ then $2z$ approaches $-\infty$, which isn't the same limit? Or something like that?

A:

Having a critical point "at infinity" makes sense when you're looking at the Riemann sphere, which you can think of as the union of $\mathbb C$ with a single point that we call $\infty$, and which is useful in complex dynamics (which includes the study of the Mandelbrot set and Julia sets). If you want to evaluate a rational function $\phi(z) = f(z)/g(z)$ whose numerator and denominator share no common factors, at a point $c\in\mathbb C$ where $g(c) = 0$, we define $\phi(c) = \infty$.
We say that a rational function $\phi(z)$ has a critical point at $\alpha\in\mathbb C\cup\{\infty\}$ if for all Möbius transformations $f(z)$ such that $\beta = f^{-1}(\alpha)\neq \infty$ and $\phi^f(\beta) = (f^{-1}\circ\phi\circ f)(\beta)\neq \infty$, $(\phi^f)'(\beta) = (f^{-1}\circ\phi\circ f)'(\beta) = 0$. One can show that this is a extension of the familiar definition of a critical point: if $\alpha\neq\infty$ and $\phi(\alpha)\neq\infty$, then $\alpha$ is a critical point of $\phi$ (according to the definition in the previous sentence) iff $\phi'(\alpha) = 0$. Note that we when refer to a derivative in this context, it is a formal derivative; for example, we are formally defining
$$
\phi'(z) = \frac{g(z)f'(z) - f(z)g'(z)}{(g(z))^2}.
$$
Now, the following lemma is quite handy:
Lemma. Let $\phi$ be a rational function and let $f$ be a Möbius transformation. Then $\forall \alpha\in\mathbb C\cup\{\infty\}$, $\alpha$ is a critical point of $\phi$ iff $\beta = f^{-1}(\alpha)$ is a critical point of $\phi^f$.
Using this lemma, we can verify Isaac's suggestion that every polynomial of degree at least two has infinity as a critical point. Consider a polynomial function $\phi(z) = a_n z^n + \cdots + a_1z + a_0$, where $n\geq 2$. Define the Möbius transformation $f(z) = 1/z$. Then $f^{-1}(\infty) = 0$, and
    \begin{align*}
 \phi^f(z) & = (f^{-1}\circ\phi\circ f)(z) \\
 & = \frac{1}{\tfrac{a_n}{z^n} + \cdots + \tfrac{a_1}{z} + a_0} \\[3pt]
 & = \frac{z^n}{a_n + \cdots + a_1z^{n-1} + a_0z^n}.
 \end{align*}
    By writing out the formal derivative $(\phi^f)'(z)$, we can check that $(\phi^f)'(0) = 0$, so clearly $0$ is a critical point of $\phi^f$. It then follows from the lemma that $\infty$ is a critical point of $\phi$.

A:

I tend to think of critical points (in a calculus sense) as places where the derivative is zero or undefined/infinite (or possibly discontinuous).

