Everybody loves the good old quadratic Mandelbrot set. As you probably know, both it and the corresponding quadratic Julia sets are defined by the iteration $f(z) = z^2 + c$.
You might expect, however, that $f(z) = az^2 + bz + c$ would give you more possibilities. However, all the books on the subject assert that this is not the case. I'm trying to get to the bottom of why this is so.
For a start, you can see that if you divide the entire thing through $a$, then the specific values taken by $f$ would change, but their relationship would not. Hence, multiplying by $a$ is only scaling and/or rotating the system. It doesn't actually change its behaviour.
But what of the linear term, $bz$? Why is that redundant?
We can ask a similar question about the cubic Mandelbrot set. A lot of people define this as $g(z) = z^3 + c$, but the definition I like is $h(z) = z^3 - 3a^2z + b$. This has two critical points (whatever that means), which yields strange, shadowy images. More interestingly, with two complex-valued parameters, the corresponding Mandelbrot set is four-dimensional!
Again, we are told that $h(z)$ is the most general formulation. (In particular, you don't need a quadratic term.) The strange formulation $-3a^2z$ rather than just $az$ seems necessary to make the parameter plane plot correctly. (It also means that the critical points are $+a$ and $-a$ exactly.)
Does anybody know why this formulation is correct? What would the general 4th order case look like?