The nth root of the power formula has n² unique valid roots (branches) if n is an integer. If n is not an integer, then depending on the values of {x, y, z}, some of the roots will not be valid.

Here is some C++ code in case you have difficulty following the math formulas below:

//Arbitrary integer roots, allows for n² branches
//this code is written for ease of understanding and is not intended to be efficient
//all unique valid roots can be found using all possible combinations of:
// ktheta=0,1,...abs(n)-1
// kphi=0,1,...abs(n)-1
triplex TriplexRoot(triplex p, int n, int ktheta, int kphi) {
 int k1=(abs(n)-(p.z<0?0:1))/4, k2=(3*abs(n)+(p.z<0?4:2))/4, dk=0;
 if(abs(n)%2==0 && kphi>k1 && kphi<k2) dk=sign(p.z)*(abs(n)%4==0?-1:1);
 double r=sqrt(sqr(p.x)+sqr(p.y)+sqr(p.z));
 double theta=(atan2(p.y,p.x)+(2*ktheta+dk)*pi)/n;
 double phi=(asin(p.z/r)+(2*kphi-dk)*pi)/n;
 return pow(r,1.0/n)*triplex(cos(theta)*cos(phi),sin(theta)*cos(phi),sin(phi));
}

Also, here is some Mathematica code in case that is easier for you to understand:
TriplexRoot[ {x_, y_, z_}, n_, ktheta_, kphi_] := Module[ {dk = If[ Mod[ n, 2] == 0 && Abs[ n] < 4kphi + If[ z < 0, 0, 1] <= 3Abs[ n], Sign[ z]If[ Mod[ n, 4] == 0, -1, 1], 0], r = Sqrt[ x^2 + y^2 + z^2],theta, phi}, theta = (ArcTan[ x, y] + (2ktheta + dk)Pi)/n; phi = (ArcSin[ z/r] + (2kphi - dk)Pi)/n; r^(1/n){Cos[ theta] Cos[ phi], Sin[ theta]Cos[ phi], Sin[ phi]}];

Have you tried defining transcendental functions by their Taylor series, such as where is a triplex, to see if that makes any kind of sense?

FWIW, the triplex numbers do not even form an alternative algebra, and there is no distributivity either.  They only have power associativity (which is no big news) .

Thank you for this definition! I think you added this to wikipedia?! (http://en.wikipedia.org/wiki/Hypercomplex_number#Three-dimensional_complex_numbers_based_on_spherical_coordinates)

But I can't see why the third dimension can be called hypercomplex.
In the normal Mandelbrot you have the simple multiplication z² in complex and only because of i² = -1 you get the mandelbrot iteration very simple; its the property that complex numbers can do a rotation and rescaling.

So I think the 3rd dimension can't be a second complex dimension, and the multiplication and result shows that a second complex number like j will not be used (and makes no sense) because the result is more than multiplying 2 hypercomplex numbers (which are sums)! In 2D it works, but try to multiply quarternions - you can't get terms with x y and z in one component. This is impossible by squaring.

I see that the 2D mandelbrot equations are still in there... But it looks like that the "triplex" is something that can't be understand in calling it hypercomplex... there must be somthing more/other...

I fixed the mistake in Wikipedia.  BTW, I think it can be called a hypercomplex number if the third axis is taken to have j^2=0 (or maybe not because unlike dual numbers, here you have to count z^2 when you compute the norm; these numbers are strange...).  You can also say that (x,-y,-z) is the conjugate.

Also, here is a closed form expression for powers of the triplex numbers based on Chebyshev polynomials Tn(x) and Un-1(x).  These give , and without explicitly computing the trigonometric functions, so they are very handy for this kind of transformation.

I initially assume that x^2+y^2+z^2 = 1. I express the angles in terms of arccos and use the Chebyshev polynomials:






From this formula, the various pieces in http://www.fractalforums.com/index.php?action=dlattach;topic=742.0;attach=436;image should be more or less identifiable.

Since (x,y,z) does not usually have unary modulus, you should divide the arguments by and multiply the results by .  Most of the divisions cancel, and what you get is:

I think I got it!

There are maybe some small problems because this is not a full definition, but there is enought defined to derive the non-trigometric equation.
The 3rd dimension j is a very complicated thing because i*j = -1 but j² is something else...

And one more thing.  I said above that (0,0,1)^2=0, but that's not true.  I was trusting the Cartesian formulas too much, but they are clearly undefined if one of the rho_xy is zero, and the simplification is not at all trivial.

In particular:

for (0,0,z1)*(0,0,z2), the result is (-z1 z2,0,0).  This is particularly important because it basically gives the required proof for the value of e^(0,0,z).  The idea is that powers alternate between (0,0,(-1)^n z^2n+1) and ((-1)^n z^2n+2,0,0), so that the series can be split to sines and cosines.

for (0,0,z1)*(x2,y2,z2), I got but this does not really make sense to me. For example it gives (0,0,2)*(0,1,1) = (0,-2,2) but it should be (0,1,1)^3 = (0,2,2) according to the trig formula.  Apparently elevation's full range must be considered between (-pi,+pi) because the (usually redundant) range matters when the elevation is multiplied by n.  So each triplex has two spherical representations, for example (1,3 pi/2, 3 pi/4) and (1,pi/2,pi/4), and bad things happen when the cartesian formulas exchange one with the other.

It may mean that we have abandoned of power associativity too (bad!), or maybe we're just complicating the formulas unnecessarily and the real underlying space is 4D, such as . I honestly have no idea, and need to work it out a bit more on paper.  Funny though that there is an exponential operator, but not a good definition of multiplication in cartesian coordinates!

As far as fractals are concerned, it's not too bad because there is a definition of power that can be used for Mandelbrot and Newton fractals at least!  Also, the results seem to vary in sign only if you change cartesian vs. trigonometric computation of powers, so it is fine if the non-trig formulas are used for drawing!

(After this detour, the third special case is trivial from commutativity).