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and coloring each complex-valued pixel c based on how long it takes to escape a circle of radius 2. The resulting image looks like this:

My own home-brew Mandelbrot Set imager now allows me to type in an arbitrary algebraic expression f(z) in terms of the complex variable z, and generates a fractal based on iterating Newton's Method to find a solution for f(z)=0.

It occurred to me that it would be interesting to graph solutions for various orbital periods in the Mandelbrot formula. For example, if I wanted to find places where the Mandelbrot formula repeated every iteration (i.e. orbital period = 1), I would solve for z² + z = z, or z² = 0. Of course, this is a trivial case, since it is obvious that the only solution is z=0.

As one might expect, a trivial formula yields a trivial image, in this case not even a fractal... just a set of concentric circles:

z2 + z = z

[Orbital Period = 1]

21−1 = 1 Solution: Re(z) Im(z) 0 0

For the orbital period of 2, we get our first non-trivial image. This time it is a fractal. The solutions are z = +1 and z = −1, which show up as the larger horizontal eyes. This is interesting, but there is no resemblance of the Mandelbrot Set (yet).

(z2 + z)2 + z = z

[Orbital Period = 2]

22−1 = 2 Solutions: Re(z) Im(z) 0 0 −1 0

Now at orbital period 3, something interesting happens: we are starting to see a definite pear shape.

Also, because of the Fundamental Theorem of Algebra, there are 2^p−1 complex solutions, where p is the orbital period, because each higher orbital period corresponds with a polynomial based on squaring the previous period's polynomial. The reason it is 2^p−1 and not 2^p is because when the orbital period is 1, there is only 1 solution, and 1 = 2⁰.

((z2 + z)2 + z)2 + z = z

[Orbital Period = 3]

23−1 = 4 Solutions: Re(z) Im(z) 0 0 −1.75487767 0 −0.12256117 ±0.74486177

(((z2 + z)2 + z)2 + z)2 + z = z

[Orbital Period = 4]

24−1 = 8 Solutions: Re(z) Im(z) 0 0 −1 0 −1.31070264 0 −1.94079981 0 0.28227139 ±0.53006062 −0.15652017 ±1.03224711

((((z2 + z)2 + z)2 + z)2 + z)2 + z = z

[Orbital Period = 5]

25−1 = 16 Solutions: Re(z) Im(z) 0 0 −1.62541373 0 −1.86078252 0 −1.98542425 0 0.37951359 ±0.33493231 0.35925922 ±0.64251374 −0.04421236 ±0.98658098 −0.19804210 ±1.10026954 −0.50434018 ±0.56276576 −1.25636793 ±0.38032096

(((((z2 + z)2 + z)2 + z)2 + z)2 + z)2 + z = z

[Orbital Period = 6]

26−1 = 32 Solutions: Re(z) Im(z) 0 0 −1 0 −1.47601464 0 −1.75487767 0 −1.77289290 0 −1.90728009 0 −1.96677322 0 −1.99637614 0 0.38900684 ±0.21585065 0.44332563 ±0.37296242 0.39653457 ±0.60418181 0.35989274 ±0.68476202 −0.01557039 ±1.02049737 −0.11341866 ±0.86056947 −0.12256117 ±0.74486177 −0.16359826 ±1.09778064 −0.21752675 ±1.11445427 −0.59689164 ±0.66298074 −1.13800067 ±0.24033240 −1.28408496 ±0.42726890

((((((z2 + z)2 + z)2 + z)2 + z)2 + z)2 + z)2 + z = z

[Orbital Period = 7]

(((((((z2 + z)2 + z)2 + z)2 + z)2 + z)2 + z)2 + z)2 + z = z

[Orbital Period = 8]