What does the Julia set of a *cubic* function look like? As a specific example, something like

$$f(z) = z^3 + i$$

or

$$f(z) = z^3-z^2+z/2-1/3.$$

# What does the Julia set of a cubic function look like?

I found an excellent resource for this here

Here's my understanding of this: for any Julia set within the family that we have been working with, $f_c(z)=z^2 + c$, the image will show rotational symmetry of 180 degrees:

In other words, if you split the figure in half and then rotate that half 180 degrees, you will see that it perfectly matches the other half.

Now, according to the website I found, functions with a degree of three should have Julia sets with rotational symmetry of 120 degrees: enough to fit three same-looking sections into the 360 degrees of the image, if that makes sense. Here's the example they gave:

Similarly, functions with a degree of four should have 90 degree rotational symmetry, a degree of five should result in rotational symmetry of 72 degrees, etc.

I'm still unsure how to determine the Julia sets for those particular examples, though, if anyone wants to add on to this.

Anyways, happy 18th birthday to me!!

@Vanderbilt That does look very cool! To all - we're still looking for a tool that generates Julia sets of things like $z^3+i$.

A little of topic, but I found it to be good information.

zn + 1 = zn3 + c

As you can see, the only difference is in the power of z. Also you will notice that all the Julia Sets of this type will have rotational symetry of order 3, instead of 2.

http://www.jamesh.id.au/fractals/mandel/Mandel3.html

Here are actual examples

https://theory.org/fracdyn/birth%5E3/