I'm trying to recreate a graph from a paper, and the function that I'm plotting involves three separate parabolic cylinder functions that all have a complex Nu value. After getting it all typed out, I end up with an empty plot. I've gone back to basics and have fiddled around with ParabolicCylinderD, and have found that I can plot it no problem with a real Nu, but as soon as I make Nu complex I end up with an empty plot. What am I doing wrong?


I'll put in the test code I used while playing with the function to see if it worked, but it's just a case of using a complex input with the function to see if it works.


That one worked totally fine, got a good plot from it.

Plot[ParabolicCylinderD[2 I,p],{p,0,50}]

This one didn't work at all, got an empty plot as a result.

  • $\begingroup$ We cannot reproduce or work on the problem without the code for a minimal example. $\endgroup$
    – Bob Hanlon
    Jul 29, 2020 at 15:16
  • $\begingroup$ I've adjusted the main question to include my tinkering example but it's literally just a case of putting a complex number in the nu value for the ParabolicCylinderD function $\endgroup$
    – PewtDmD
    Jul 29, 2020 at 15:22
  • 1
    $\begingroup$ @PewtDmD Plot will not show complex numbers. You need to take real Re and imaginary Im separately, or if you only care about the norm, then use Abs $\endgroup$
    – flinty
    Jul 29, 2020 at 15:34
  • 2
    $\begingroup$ Depending on your version of Mathematica, one or more of the following will work: ReImPlot[ParabolicCylinderD[2 I, p], {p, 0, 8}, PlotRange -> All, PlotLegends -> Placed["ReIm", {0.5, 0.5}]] or Plot[Evaluate@ReIm@ParabolicCylinderD[2 I, p], {p, 0, 8}, PlotRange -> All, PlotLegends -> Placed[{Re, Im}, {0.5, 0.5}]] or Plot[Evaluate[#[ParabolicCylinderD[2 I, p]] & /@ {Re, Im}], {p, 0, 8}, PlotRange -> All, PlotLegends -> Placed[{Re, Im}, {0.5, 0.5}]] $\endgroup$
    – Bob Hanlon
    Jul 29, 2020 at 15:44
  • $\begingroup$ Perhaps if you had added a link to the paper you speak of, it would be easier to see what the function to be plotted is supposed to be. $\endgroup$ Sep 25, 2020 at 0:44

1 Answer 1


Yes certainly. If you have a fixed complex $\nu$ then use ComplexPlot not Plot like so:

ComplexPlot[ParabolicCylinderD[0.3 + 0.2 I, z], {z, -1 - I, 1 + I}]

enter image description here

... or if you still want to use Plot, then take the real and imaginary parts separately:

Plot[{Re[ParabolicCylinderD[0.3 + 0.2 I, x]], 
  Im[ParabolicCylinderD[0.3 + 0.2 I, x]]}, {x, -3, 3}, 
 PlotStyle -> {Directive[Red], Directive[Blue]}]

enter image description here

For $\nu=2\mathbf{i}$ we have real and imaginary parts quickly decaying towards zero:

  Re[ParabolicCylinderD[2 I, x]],
  Im[ParabolicCylinderD[2 I, x]], 
  Abs[ParabolicCylinderD[2 I, x]]}, {x, 0, 10}, 
 PlotStyle -> {Directive[Red], Directive[Blue], Directive[Thick, Darker@Green]}, 
 PlotRange -> All, 
 PlotLegends -> {"Re", "Im", "Abs"}]

enter image description here

  • $\begingroup$ The result of ComplexPlot3D[ParabolicCylinderD[0.3 + 0.2 I, z], {z, -1 - I, 1 + I}] is more informative. $\endgroup$
    – user64494
    Jul 29, 2020 at 15:38
  • 1
    $\begingroup$ Also note you can do: FunctionExpand[Abs[ParabolicCylinderD[2 I, x]]], FunctionExpand[Re[ParabolicCylinderD[2 I, x]]], and FunctionExpand[Im[ParabolicCylinderD[2 I, x]]] to get forms in exponentials, cosines, and HermiteH functions. $\endgroup$
    – flinty
    Jul 29, 2020 at 15:46

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