# Is it possible to have a complex nu in ParabolicCylinderD?

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?

Edit:

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.

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


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.

• We cannot reproduce or work on the problem without the code for a minimal example. – Bob Hanlon Jul 29 '20 at 15:16
• 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 – PewtDmD Jul 29 '20 at 15:22
• @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 – flinty Jul 29 '20 at 15:34
• 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}]] – Bob Hanlon Jul 29 '20 at 15:44
• 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. – J. M.'s torpor Sep 25 '20 at 0:44

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

Plot[{
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"}] • The result of ComplexPlot3D[ParabolicCylinderD[0.3 + 0.2 I, z], {z, -1 - I, 1 + I}] is more informative. – user64494 Jul 29 '20 at 15:38
• 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. – flinty Jul 29 '20 at 15:46