# How to plot this set in the complex plane?

I'd like to plot the set

s=(z - I Im[z] < 0 && -I z + I Re[z] ==
0) || (-1 + Cos[(9 Arg[z])/2] Im[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) +
2 Cos[(9 Arg[z])/2] Im[z]^2 Re[z]^2 (Im[z]^2 + Re[z]^2)^(1/4) +
Cos[(9 Arg[z])/2] Re[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) < 0 &&
Im[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] +
2 Im[z]^2 Re[z]^2 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] +
Re[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] == 0)


Both

ComplexContourPlot[s, {z, -10 - 10*I, 10 + 10*I}]


and

ComplexRegionPlot[(s,{z,-10-10*I,10+10*I}]


produce empty plots. I know s is not empty as

N[s /. z -> (-1)^(4/9)]


True

shows. I think its dimension equals one. The question is inspired by that question.

Addition. Making use of the Domen's comment, I replace inequalities by Booles,

ComplexContourPlot[Boole[z - I Im[z] < 0]*(-I z + I Re[z] == 0) ||
Boole[-1 + Cos[(9 Arg[z])/2] Im[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) +
2 Cos[(9 Arg[z])/2] Im[z]^2 Re[z]^2 (Im[z]^2 + Re[z]^2)^(1/4) +
Cos[(9 Arg[z])/2] Re[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) <
0]*(Im[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] +
2 Im[z]^2 Re[z]^2 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] +
Re[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] == 0), {z,  10}]


, but the result is doubtful to me.

Addition 2. The improved approach

ComplexContourPlot[Boole[z - I Im[z] < 0] == 1 && (-I z + I Re[z] == 0) ||
Boole[-1 + Cos[(9 Arg[z])/2] Im[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) +
2 Cos[(9 Arg[z])/2] Im[z]^2 Re[z]^2 (Im[z]^2 + Re[z]^2)^(1/4) +
Cos[(9 Arg[z])/2] Re[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) < 0] ==
1 && (Im[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] +
2 Im[z]^2 Re[z]^2 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] +
Re[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] == 0), {z,  10}]


results in empty plot. PlotPoints -> 300 does not help.

• FindInstance[s && Im[z] != 0, z, Complexes] fails for me in 13.3 on Windows 10. Commented Aug 30, 2023 at 18:15
• 1. You cannot use inequations for ComplexContourPlot; it can only accept a function (or equation). 2. Please see "Possible Issues" in the documentation for ComplexRegionPlot: It cannot draw one-dimensional regions (which yours is). Commented Aug 30, 2023 at 18:46
• @Domen. Thank you for your valuable comment. Commented Aug 30, 2023 at 19:01

• ComplexRegionPlot in Mathematica seems can't handle 2D region and 1D lines simultaneously.

• Here we try to use ComplexContourPlot to plot the 1D lines and use RegionFunction to limit the 2D region.

Clear[plot1, plot2];
plot1 = ComplexContourPlot[-I z + I Re[z] == 0, {z, 10},
RegionFunction -> Function[{z, f}, z - I Im[z] < 0],
ContourStyle -> Green]; plot2 =
ComplexContourPlot[
Im[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] +
2 Im[z]^2 Re[z]^2 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] +
Re[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] == 0, {z, 10},
RegionFunction ->
Function[{z,
f}, -1 + Cos[(9 Arg[z])/2] Im[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) +
2 Cos[(9 Arg[z])/2] Im[z]^2 Re[z]^2 (Im[z]^2 + Re[z]^2)^(1/4) +
Cos[(9 Arg[z])/2] Re[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) < 0],
ContourStyle -> Blue];
Show[plot1, plot2]


• The above method can also illustrate by using MeshFunctions in ComplexRegionPlot. Here we only illustrate the second part.
ComplexRegionPlot[(-1 +
Cos[(9 Arg[z])/2] Im[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) +
2 Cos[(9 Arg[z])/2] Im[z]^2 Re[z]^2 (Im[z]^2 + Re[z]^2)^(1/4) +
Cos[(9 Arg[z])/2] Re[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) < 0), {z, 10},
PlotPoints -> 200, MaxRecursion -> 4,
MeshFunctions ->
Function[{z},
Im[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] +
2 Im[z]^2 Re[z]^2 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2] +
Re[z]^4 (Im[z]^2 + Re[z]^2)^(1/4) Sin[(9 Arg[z])/2]],
Mesh -> {{0}}, MeshStyle -> Blue, PlotStyle -> Opacity[.1],
BoundaryStyle -> Opacity[.1]]