# How can I generate this “domain coloring” plot?

I found this plot on Wikipedia: Domain coloring of $\sin(z)$ over $(-\pi,\pi)$ on $x$ and $y$ axes. Brightness indicates absolute magnitude, saturation represents imaginary and real magnitude.

Despite following the link and reading the page nothing I have tried is giving me the result shown. How should this be done?

• And what have you tried? – Ajasja Jun 23 '12 at 11:38
• @Ajasja that's fair, but I ask you to cut me some slack: I've posted 395 answers to this site so I'm not unwilling to exert myself. In this case I was having a mental block so nothing I tried was worth sharing. – Mr.Wizard Jun 23 '12 at 17:41
• Is this homework? – Dr. belisarius Jun 24 '12 at 4:16
• @belisarius lol :-) (I'm assuming that's your quirky humor again.) – Mr.Wizard Jun 24 '12 at 6:00
• It's is noteworthy that Claudio Rocchini provided C/C++ code for en.wikipedia.org/wiki/File:Color_complex_plot.jpg on that same page – Tobias Kienzler Aug 16 '13 at 6:41

Building on Heike's ColorFunction, I came up with this: The white bits are the trickiest - you need to make sure the brightness is high where the saturation is low, otherwise the black lines appear on top of the white ones.

The code is below. The functions defined are:

• complexGrid[max,n] simply generates an $n\times n$ grid of complex numbers ranging from $-max$ to $+max$ in both axes.

• complexHSB[Z] takes an array $Z$ of complex numbers and returns an array of $\{h,s,b\}$ values. I've tweaked the colour functions slightly. The initial $\{h,s,b\}$ values are calculated using Heike's formulas, except I don't square $s$. The brightness is then adjusted so that it is high when the saturation is low. The formula is almost the same as $b2=\max (1-s,b)$ but written in a way that makes it Listable.

• domainImage[func,max,n] calls the previous two functions to create an image. func is the function to be plotted. The image is generated at twice the desired size and then resized back down to provide a degree of antialiasing.

• domainPlot[func,max,n] is the end user function which embeds the
image in a graphics frame.

complexGrid = Compile[{{max, _Real}, {n, _Integer}}, Block[{r},
r = Range[-max, max, 2 max/(n - 1)];
Outer[Plus, -I r, r]]];

complexHSB = Compile[{{Z, _Complex, 2}}, Block[{h, s, b, b2},
h = Arg[Z]/(2 Pi);
s = Abs[Sin[2 Pi Abs[Z]]];
b = Sqrt[Sqrt[Abs[Sin[2 Pi Im[Z]] Sin[2 Pi Re[Z]]]]];
b2 = 0.5 ((1 - s) + b + Sqrt[(1 - s - b)^2 + 0.01]);
Transpose[{h, Sqrt[s], b2}, {3, 1, 2}]]];

domainImage[func_, max_, n_] := ImageResize[ColorConvert[
Image[complexHSB@func@complexGrid[max, 2 n], ColorSpace -> "HSB"],
"RGB"], n, Resampling -> "Gaussian"];

domainPlot[func_: Identity, max_: Pi, n_: 500] :=
Graphics[{}, Frame -> True, PlotRange -> max, RotateLabel -> False,
FrameLabel -> {"Re[z]", "Im[z]",
"Domain Colouring of " <> ToString@StandardForm@func@"z"},
BaseStyle -> {FontFamily -> "Calibri", 12},
Prolog -> Inset[domainImage[func, max, n], {0, 0}, {Center, Center}, 2 max]];

domainPlot[Sin, Pi]


Other examples follow:

It's informative to plot the untransformed complex plane to understand what the colours indicate:

domainPlot[] A simple example:

domainPlot[Sqrt] Plotting a pure function:

domainPlot[(# + 2 I)/(# - 1) &] I think this one is very pretty:

domainPlot[Log] • +1, brilliant! In addition to just being lovely images, I love how the poles and branch cuts are clearly visible. – rcollyer Jun 27 '12 at 14:56
• Incidentally, I posted this on Facebook. The response I got: "Holy crap, mathematica keeps getting more and more awesome." – rcollyer Jun 27 '12 at 15:53
• Thanks for that, I've updated the code to work in version 7. – Simon Woods Jun 28 '12 at 9:43
• @rcollyer, that's great! Your friend has excellent taste. My friends would just think I was weird if I posted domain colouring plots on Facebook :-) – Simon Woods Jun 28 '12 at 9:44
• On Mathematica V.9, the complexGrid[] function was wrong for me, the Outer[] was not giving the expected result (it puts z with Re[z]<0 and Im[z]<0 in the upper left quadrant). See e.g. the following code: Outer[Plus, I {-1, 0, 1}, {-1, 0, 1}] // MatrixForm I replaced it with Outer[Plus, r, I r // Reverse]\[Transpose] which works for me. – user5665 Jan 31 '13 at 13:19

Not as pretty as the one in the original post, but it's getting in the right direction I think:

RegionPlot[True,
{x, -Pi, Pi}, {y, -Pi, Pi},
ColorFunction -> (Hue[Rescale[Arg[Sin[#1 + I #2]], {-Pi, Pi}],
Sin[2 Pi Abs[Sin[#1 + I #2]]]^2,
Abs@(Sin[Pi Re[Sin[#1 + I #2]]] Sin[Pi Im[Sin[#1 + I #2]]])^(1/
4), 1] &),
ColorFunctionScaling -> False, PlotPoints -> 200] It seems that the hue of the colour function is a function of Arg[Sin[z]], saturation is a function of Abs[Sin[z]] and the brightness is related to Re[Sin[z]] and Im[Sin[z]].

• +1 Very close. What did you wanted to say about the brightness? – Matariki Jun 23 '12 at 12:09
• ...and if you use the color scheme in Thaller's package along with Heike's idea, you get this, whose coloring is a wee bit closer to the one in the OP. – J. M.'s technical difficulties Jun 23 '12 at 12:28

This is a good way :

DensityPlot[ Rescale[ Arg[Sin[-x - I y]], {-Pi, Pi}], {x, -Pi, Pi}, {y, -Pi, Pi},
MeshFunctions -> Function @@@ {{{x, y, z}, Re[Sin[x + I y]]},
{{x, y, z}, Im[Sin[x + I y]]},
{{x, y, z}, Abs[Sin[x + I y]]}},
MeshStyle -> {Directive[Opacity[0.8], Thickness[0.001]],
Directive[Opacity[0.7], Thickness[0.001]],
Directive[White, Opacity[0.3], Thickness[0.006]]},
ColorFunction -> Hue, Mesh -> 50, Exclusions -> None, PlotPoints -> 100] Another ways to tackle the problem, which apprears promising.

ContourPlot[ Evaluate @ {Table[Re @ Sin[x + I y] == 1/2 k, {k, -25, 25}],
Table[Im @ Sin[x + I y] == 1/2 k, {k, -25, 25}]},
{x, -Pi, Pi}, {y, -Pi, Pi}, PlotPoints -> 100, MaxRecursion -> 5] and

RegionPlot[ Evaluate @ {Table[1/2 (k + 1) > Re @ Sin[x + I y] > 1/2 k, {k, -25, 25}],
Table[1/2 (k + 1) > Im @ Sin[x + I y] > 1/2 k, {k, -25, 25}]},
{x, -Pi, Pi}, {y, -Pi, Pi}, PlotPoints -> 50, MaxRecursion -> 4,
ColorFunction -> Function[{x, y}, Hue[Re@Sin[x + I y]]]] These plots seem to be good points for further playing around to get better solutions.

I already mentioned Bernd Thaller's package GraphicsComplexPlot in the comments; if one blends the ideas from Artes's and Heike's answers, and then use the function $ComplexToColorMap[] from Thaller's package (I won't include it here; again, see the package for that), we get this: Needs["GraphicsComplexPlot"] (* Thaller's package; get it yourself *) f1 = RegionPlot[True, {x, -Pi, Pi}, {y, -Pi, Pi}, ColorFunction -> ($ComplexToColorMap[Abs[Sin[#1 + I #2]],
Arg[Sin[#1 + I #2]], {Pi, 1/10, 1, 1/10, 1}] &),
ColorFunctionScaling -> False, PlotPoints -> 200];

f2 = ContourPlot[
Evaluate@{Table[Re@Sin[x + I y] == 1/2 k, {k, -25, 25}],
Table[Im@Sin[x + I y] == 1/2 k, {k, -25, 25}]}, {x, -Pi,
Pi}, {y, -Pi, Pi}, PlotPoints -> 100, ContourStyle -> Gray];

f3 = ContourPlot[
Evaluate@Table[Abs@Sin[x + I y] == 1/2 k, {k, -25, 25}], {x, -Pi,
Pi}, {y, -Pi, Pi}, PlotPoints -> 100, ContourStyle -> White,
MaxRecursion -> 5];

Show[f1, f2, f3]


The $ComplexToColorMap[] function could probably be optimized a fair bit for new Mathematica, but I won't get into that for now. One might also consider tweaking the Opacity[] of the contour lines for the absolute value as well, but I'll leave that as an experiment for the reader. Another thing you can try: RegionPlot[True, {x, -Pi, Pi}, {y, -Pi, Pi}, ColorFunction -> ($ComplexToColorMap[Abs[Sin[#1 + I #2]],
Arg[Sin[#1 + I #2]], {Pi, 1/50, 1, 1/50, 1}] &),
ColorFunctionScaling -> False, Mesh -> 51,
MeshFunctions -> {Re[Sin[#1 + I #2]] &, Im[Sin[#1 + I #2]] &},
MeshStyle -> Gray, PlotPoints -> 95]


With Mathematica 12.0, there's now a ComplexPlot function that replaces user made solutions. As with other Plot functions, it allows us to specify a ColorFunction option to manipulate how to color the plot. This particular coloring is implemented natively in the "CyclicReImLogAbs" option.

So the modern equivalent is

ComplexPlot[Sin[z], {z, -Pi - Pi I, Pi + Pi I},
ColorFunction -> "CyclicReImLogAbs", Frame -> False]
` • complements user made solutions* – C. E. Apr 16 '19 at 20:47