2

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}]

enter image description here

, 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.

user64494
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  • FindInstance[s && Im[z] != 0, z, Complexes] fails for me in 13.3 on Windows 10. – user64494 Aug 30 '23 at 18:15
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  • 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).
  • – Domen Aug 30 '23 at 18:46
  • @Domen. Thank you for your valuable comment. – user64494 Aug 30 '23 at 19:01