How to plot "Re-Im" 3d plot with `ComplexPlot3D` (`MeshFunctions -> {Re[#2] &, Im[#2] &}` should work, but doesn't)

Question

By default the

f[z_] = ((z^2 - 1)*((z - 2 - I)^2)) / (z^2 + 2 + 2*I) 
ComplexPlot3D[f[z], {z, 10}]

will output AbsArg 3d plot (because by default the MeshFunctions -> {Abs[#2] &, Arg[#2] &})

But I want ReIm 3d plot like here https://mathematica.stackexchange.com/a/125633/86077

But there should be easier way

But when I try MeshFunctions -> {Re[#2] &, Im[#2] &} it doesnt work, it still shows AbsArg plot

WHY?
How MeshFunctions work? how the output is used?

What are you want to plot ? Abs@f[z] or ReIm@f[z]? ComplexPlot3D only plot Abs@f[z]. — cvgmt, Apr 15 '22 at 22:22
If f is a function from x + y i to re + im i them i want x y as bottom axis , re as vertical axis and im as temperature — srghma, Apr 16 '22 at 07:27
ComplexPlot3D plots abs as vertical axis and arg as temperature — srghma, Apr 16 '22 at 07:28

cvgmt · Accepted Answer · 2022-04-16T08:04:21.543

3

f[z_] = ((z^2 - 1)*((z - 2 - I)^2))/(z^2 + 2 + 2*I);
Block[{z = x + I*y}, 
 Plot3D[Re@f[z], {x, -5, 5}, {y, -5, 5}, 
  ColorFunction -> 
   Function[{x, y}, ColorData["TemperatureMap"][Im@f[z]]], 
  ColorFunctionScaling -> True]]

We can add Mesh and the same time.

f[z_] = ((z^2 - 1)*((z - 2 - I)^2))/(z^2 + 2 + 2*I);
Block[{z = x + I*y}, 
 Plot3D[Re@f[z], {x, -5, 5}, {y, -5, 5}, 
  ColorFunction -> 
   Function[{x, y}, ColorData["TemperatureMap"][Im@f[z]]], 
  MeshFunctions -> Function[{x, y}, Im@f[z]], 
  ColorFunctionScaling -> True]]

edited Apr 16 '22 at 08:04

answered Apr 16 '22 at 07:59

cvgmt

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But the idea of this question is how to do this with ComplexPlot3D? – srghma Apr 16 '22 at 08:26
1

@srghma It is impossible to do this by ComplexPlot3D since ComplexPlot3D can only plot Abs@f[z]. – cvgmt Apr 16 '22 at 08:30
Ah, ok, also now i understand (after seeing second example) what MeshFunctions are doing – srghma Apr 16 '22 at 08:34

How to plot "Re-Im" 3d plot with `ComplexPlot3D` (`MeshFunctions -> {Re[#2] &, Im[#2] &}` should work, but doesn't)

1 Answers1