Mathematica can plot the complex function using abs and arg, but I want to plot a complex function
with colors encoding the imaginary part. I cannot figure it out. Thanks in advance!
Daniel,
The problem is syntax:
You entered
ColorFunction -> Function[{x, y, z}, Hue[Im[z^2 + 1]]]
In this function, the x is the real part of z, the y is the Imaginary part of z, BUT z is not your original z its the magnitude of Re[z^2 + 1]. Using z in two places like that is confusing and leads to errors.
You can fix it with:
fun = x + y I;
Plot3D[Re[fun^2 + 1], {x, -2, 2}, {y, -2, 2},
ColorFunction -> Function[{x, y, z}, Hue[Im[fun^2 + 1]]],
PlotLegends -> BarLegend[Hue[-4, 4]]]
to get
Regards,
Neil
Daniel,
First, to get your bar legend, you need to make a change in syntax:
Plot3D[Re[fun^2 + 1], {x, -2, 2}, {y, -2, 2},
ColorFunction -> Function[{x, y, z}, Hue[Im[(x + y I)^2 + 1]]],
PlotLegends -> BarLegend[{Hue, {-4, 4}}]]
As to the Color function, the problem is the default behavior is ColorFunctionScaling->True which scales your x,y, and z to the range 0 to 1. This means that your calculation of Im[fun^2+1] is no longer correct. You need to use ColorFunctionScaling->False, and then you need to scale the values of the argument to Hue to make sense (between 0 and 1). This means you need an offset and scale to Hue to get your function values between 0 and 1. You can automate this easily but the basic plot should look something like this:
Plot3D[Re[fun^2 + 1], {x, -2, 2}, {y, -2, 2},
ColorFunction ->
Function[{x, y, z}, Hue[(Im[(x + y*I)^2 + 1] + 8)/16]],
ColorFunctionScaling -> False, PlotLegends -> BarLegend[Hue]]
The legend has the wrong values but you can fix that by creating a function to give to BarLegend that scales the values so the bar is correct.
You get the following:
Regards,
Neil
Thank you, Neil.
I didn’t expect it to be so complicated to use Mathematica to visualize the complex function encoding the imaginary part.
Thank you.
Glad to help, I’m also glad you caught my error in the first post – I did not consider the scaling problem (although I have been burned by that multiple times in the past!) Whenever I used a custom color function I frequently had to alter the ColorFunctionScaling to get proper results.
Regards,
Neil
Dear Neil,
I want to know why you use Im[(x + yI)^2 + 1] + 8)/16 to scale the value to [0,1]?
why not use Im[(x + yI)^2 + 1] /8 directly?
thank you!
By the way, could you please provide the code for scaling the bar? I have a tried, but failed.
Thank you!
Daniel,
Sorry I missed your request.
I want to know why you use Im[(x + yI)^2 + 1] + 8)/16 to scale the value to [0,1]? why not use Im[(x + yI)^2 + 1] /8 directly?
The Hue function ranges from 0 to 1 but your data ranges from -8 to 8 so you need to both scale and offset the function to get a full range of hues in the plot. Dividing by 8 gives you a range of -1 to 1. Dividing by 16 gives you a range of -0.5 to 0.5 and we shift it by 0.5 (8/16) to go from 0 to 1.
By the way, could you please provide the code for scaling the bar?
Plot3D[Re[fun^2 + 1], {x, -2, 2}, {y, -2, 2},
ColorFunction -> Function[{x, y, z}, myHue[Im[(x + y*I)^2 + 1]]],
ColorFunctionScaling -> False,
PlotLegends -> BarLegend[{myHue[#] &, {-8, 8}}]]
Regards,
Neil