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I'm making a Mandelbulb in Processing and I have a version that works by using rotation with Spherical Coordinates and wanted to make one that works with triplex multiplication. So far I only tried doing a triplex version of z = z^2+c (instead of Mandelbulb z = z^8+c) but the result was a slightly 3D Mandelbrot (the sides have a round bulge). I apologize if there is a better name for this, but it is not the same as my result for the rotation method to the second power. I'll post images of both at the bottom.

I'm using this as reference: http://www.fractalforums.com/theory/triplex-algebra/

Specifically I thought I'd use this part:

{x1, y1, z1}*{x2, y2, z2} = {(x1*x2-y1*y2)*(1-(z1*z2)/(p1*p2)), (x2*y1+x1*y2)*(1-(z1*z2)/(p1*p2)), p1*z2+p2*z1}

p1 = sqrt(x1*x1+y1*y1)
p2 = sqrt(x2*x2+y2*y2)

I made the assumption that if (x1, y1, z1) is z then (x1, y1, z1) is equal to (x2, y2, z2) because I'm trying to square z. This is the section of my code that is supposed to be doing the z = z^2+c math:


Still working on rendering the points better... hopefully it is still obvious what fractals these are.

Result from rotation method at z = z^2+c and max iterations of 10:


Result from triplex method at z = z^2+c and max iterations of 10:

your assumption is correct, but the multiplication is developed from the exponentiation, hence you just need to implement the exponentiation, which is the first
and i would suggest you include the exponentiation because of easier playing with other exponents

after you got this you just need the "+" operation which is standard vector math

and just remember, if z=0 it is standard complex algebra

yes, this is the exponentiation or power formula, a standard mandelbrot is retrieved when using "2" as "power"
both 3d fractals look rather disappointing in the power2 mode, and the original mandelbulb is a power8 mandelbrot
fractal
 
a good test if your formula is correct is to just render the z=0 plane, which should yield you then a mandelbrot,
then you just need to inspect the z= part of the formula to make it correct

both formulas yield mandelbrot shape in z=0 plane!

side note:
just to let you know what you are doing: the original bailout is sqrt(z)<4, you let the sqrt away, but it would be better
to use z<16 as bailout, because sqrt(16)=4, this way you can ommit the sqrt in the bailout test, or you know what
you where doing, then the 4 is fine because you experimented already with the bailout limit

the non trig is supposed to make the same image, just check your formula carefully, usually its just a sign +/- error, or something similar easy to overlook

I finally got it! It smokes my exponentiation code speed wise



I found bugman's website that had a similar equation to the one I was using: http://bugman123.com/Hypercomplex/

Next step is to render it better and then do the triplex method for each increasing power up to 8. Thank you in particular to cKleinhuis for your patience helping out.