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(http://u5789.direct.atpic.com/24801/0/1201302/1024.jpg) (http://pic.atpic.com/1201302/1024)

This Julia set is typical of a class of this function's Julia sets that have a quadratic Julia set border (with the Julia seed not quite constant) and filigrees, lakes, or other internal structures. The outside points diverge. Inside points mostly converge to zero. Points reached from infinity by crossing an odd number of lines go to zero, others to infinity.

The image uses the distance estimator and the following gradient:

• black further than one unit from all basin boundary points;
• Shading logarithmically to white 1/10,000 of a unit from any basin boundary point; and
• white within 1/10,000 of a unit of any basin boundary point.

A "unit" is not exactly a pixel, nor a constant distance, because the Herman Ring distance estimator is somewhat less accurate than the quadratic one for some reason. The dynamic plane distance estimation tends to underestimate the distance significantly for points near zero.
This is the cause of the "bloom" effect near the lower-central area.

Still, to a fair approximation, the image shows the Julia set itself in white on black.

This Julia set does not contain any actual Herman rings. The only attractors are the two obligatory ones, zero and infinity, for this family of mappings.

Freely redistributable and usable subject to the Creative Commons Attribution license, version 3.0.

Detailed stats:
Name: Julia's Curls
Date: January 29, 2009
Fractal: Herman Ring Julia set
Location: angle parameter = 0.21803398, c = 0.2291666667 + 3.291666667i
Depth: Very Shallow
Min Iterations: 1
Max Iterations: 127
Layers: 1
Anti-aliasing: 3x3, threshold 0.10, depth 1
Preparation time: 1 minute
Calculation time: 5 minutes (2GHz dual-core Athlon XP)

On the subject of distance estimation:

http://MakinMagic.deviantart.com/art/Fixed-Newton-DE-104288692

Uses mmf.ulb:MMF Newton Distance Estimator as the outside colouring plug-in.
I'm quite pleased with the algorithm, it just has the static points at the roots and root echoes remaining - I think they could probably be removed by somehow incorporating the second derivative into the calculations but I haven't had the time to play with that idea yet :)