0
$\begingroup$
ContourPlot3D[{Im[-1.141088662*I - (.2852721655*Sqrt[8]*Sqrt[2])/ Tan[(2*Sqrt[2]*(0.6510416668*I*Sqrt[8]*x + (.2260561342*I)*Sqrt[8]*t^.5 + 1))]]},
  {x, -10, 10}, {t, -10, 10}, {y, -10, 10}, {z, -10, 10}]
$\endgroup$
1
  • 4
    $\begingroup$ if you look at help it says ContourPlot3D has the form ContourPlot3D[...,{},{},{}] but your command has the form ContourPlot3D[...,{},{},{},{}] so you have an extra {} there. There should be only 3 {} and not 4 {} $\endgroup$
    – Nasser
    Commented Mar 31, 2020 at 20:18

1 Answer 1

2
$\begingroup$

Your function does not depend on $y$ and $z$:

ContourPlot[Im[-1.141088662*I - (.2852721655*Sqrt[8]*Sqrt[2])/
  Tan[(2*Sqrt[2]*(0.6510416668*I*Sqrt[8]*x + (.2260561342*I)*Sqrt[8]*Sqrt[t] + 1))]],
  {x, -10, 10}, {t, -10, 10}]

enter image description here

$\endgroup$
1
  • $\begingroup$ For a cleaner plot: ContourPlot[ Evaluate[Im[-1.141088662*I - (.2852721655*Sqrt[8]*Sqrt[2])/ Tan[(2*Sqrt[ 2]*(0.6510416668*I*Sqrt[8]*x + (.2260561342*I)*Sqrt[8]* Sqrt[t] + 1))]] // Rationalize[#, 0] & // Simplify], {x, -10, 10}, {t, -10, 10}, PlotPoints -> 150, WorkingPrecision -> 15] $\endgroup$
    – Bob Hanlon
    Commented Apr 1, 2020 at 1:52

Not the answer you're looking for? Browse other questions tagged or ask your own question.