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So I have to generate a few different plots with z, where z is a complete number...

z[x_, y_] := x + y*I
F[z_] := (25*Pi*z*I)/(1 + 10*Pi*z*I)

First, I need to graph u(x,y) and v(x,y) which are the real and complex parts of F(z), respectively. My confusion lies in how to make it so the plot is in terms of x and y as opposed to z and F(z)...

When trying to graph the code below, nothing shows up...

 ContourPlot[{(z[x, y])*(Conjugate[z[x, y]] + 2) == 3}, {x, -100, 
  100}, {y, -100, 100}, PlotTheme -> "Scientific"]

Thank you!


marked as duplicate by Jens, bbgodfrey, Kuba, Bob Hanlon, Dr. belisarius plotting Feb 28 '15 at 3:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Welcome to Mathematica.SE! I suggest that: 1) You take the introductory Tour now! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! 3) As you receive help, try to give it too, by answering questions in your area of expertise. $\endgroup$ – bbgodfrey Feb 27 '15 at 21:18
  • $\begingroup$ Check the documentation for SetDelayed to better understand the use of :=. In that context, replace F[z] by F[z_] in your second line of code. $\endgroup$ – bbgodfrey Feb 27 '15 at 21:21
  • $\begingroup$ You defined a function called z[x_,y_] that takes 2 argument. But then on next line below it, you wrote z with no arguments. ?? $\endgroup$ – Nasser Feb 27 '15 at 21:24

Starting with your code, corrected as in my comment,

z[x_, y_] := x + y*I 
F[z_] := (25*Pi*z*I)/(1 + 10*Pi*z*I)

you can plot the imaginary part of F as follow

ContourPlot[Im[F[z[x, y]]], {x, -.2, .2}, {y, -.2, .2},  PlotRange -> All, 
  Contours -> Range[-5, 5, .5], ContourLabels -> True]

enter image description here

and similarly for other quantities. Many different types of plots are available.

  • $\begingroup$ Ah! So you can have both of them with added underscore - I thought this would mess up redefining z when x and y are being redefined. I read the definitions - makes much more sense now. Thank you! $\endgroup$ – Jobelle Holcombe Feb 27 '15 at 23:00
  • $\begingroup$ When I've tried this however nothing shows up on the plot ContourPlot[{(z[x, y])*(Conjugate[z[x, y]] + 2) == 3}, {x, -100, 100}, {y, -100, 100}, PlotTheme -> "Scientific"] $\endgroup$ – Jobelle Holcombe Feb 28 '15 at 2:42
  • $\begingroup$ Remember that ContourPlot cannot handle complex numbers. ContourPlot[Re[z[x, y] (Conjugate[z[x, y]] + 2)] == 3, {x, -100, 100}, {y, -100, 100}] plots a small circle. $\endgroup$ – bbgodfrey Feb 28 '15 at 12:58

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