Pretty pictures.

[andydude]

Apart from that the pictures look good. What type of plot is it (conformal or contour?) and how is it encoded in colors? How did you compute the 4-roots?

Hue[h, s, b] is Mathematica notation for Hue-Saturation-Brightness encoding of colors, I found this function here and used it in my own code:

Code:

PrettyHue[Indeterminate]
    := Hue[1, 0, 0];
PrettyHue[z_] :=
    Hue[N[Mod[Arg[z], 2Pi]/(2Pi)],
      1/(1 + 0.3 Log[Abs[z] + 1]),
      1 - 1/(1.1 + 5Log[Abs[z] + 1])];

The idea is that white = infinity, black = 0, and red = positive real, cyan/blue = negative real, and all other colors represent the angle of the complex number. Roughly speaking, if \( z = r e^{i \theta} \), then \( \theta \) determines the hue, and \( r \) determines the brightness. I think it is a brilliant way to show complex functions. Much more "smooth" than a contour plot, in my opinion.

Sorry I should have said more about each function. So back to the functions.

TetraRoot00 is \( f(z) = z^{1/z} \), sorry for the silly name.
TetraRoot2 is \( f(z) = \ln(z)/W(\ln(z)) \), which requires a good CAS.

TetraPow2 is \( f(z) = z^z \), which is pretty easy to compute.
TetraPow3 is \( f(z) = z^{z^z} \), which is also easy to compute, but slow.
TetraPow4 is \( f(z) = z^{z^{z^z}} \), which was very slow (attached below).

I am currently in the process of trying to work out the complex structure of TetraRoot3, or \( f(z) \) such that \( {}^{3}(f(z)) = z \). Its not as "simple" as TetraRoot2, because instead of branch cuts on the real axis, the branch cuts are away from the real axis, I believe you can see this where the zeros are in TetraPow3' and if the derivative is zero, then the inverse function (TetraRoot3) should have a singularity at that point, right?

This is TetraPow3':
   

This is TetraPow4: