Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have this function:

$$z(x,y) = \left| \frac{ \frac{1}{3x +iy} -2x}{iy + \frac{1}{x}} \right| $$

I want the contour plot of $\frac{\partial z}{\partial x}$ with axes $(x,y)$. Tried this code, but didn't work.

z = Abs[ ( (1/3 x + I y) - 2 x )/ (I y + 1/x) ]
ContourPlot[D[z[x, y],x], {x, -2, 2}, {y, -2, 2}]

This seems rather straightforward, so I'm not sure what I'm missing.

share|improve this question
One issue is that you defined z as an expression but you're using it as a function. Another issue might be the derivative of Abs; try using ComplexExpand.` – b.gatessucks Aug 5 '14 at 9:27
How do I fix that? – user44840 Aug 5 '14 at 9:33
You can do z[x_,y_]=ComplexExpand[..., TargetFunctions->{Re, Im}]. – b.gatessucks Aug 5 '14 at 9:36
Following what @b.gatessucks said you have this. – Öskå Aug 5 '14 at 9:48
@Öskå Ha, you beat me to it :). – Teake Nutma Aug 5 '14 at 9:52
up vote 5 down vote accepted

As @b.gatessucks said in the comments, there are two issues with your code. First, you'll need to define z as a function with SetDelayed, and also add in ComplexExpand:

z[x_, y_] := ComplexExpand @ Abs[((1/3 x + I y) - 2 x)/(I y + 1/x)];

Sqrt[(25 x^2)/9 + y^2] / Sqrt[x^-2 + y^2]

Additionally, ContourPlot holds its arguments (i.e. it doesn't evaluate them), so you'll also need to throw in an Evaluate to successfully plot the contours:

ContourPlot[Evaluate @ D[z[x, y], x], {x, -2, 2}, {y, -2, 2}]

Mathematica graphics

share|improve this answer
Since b.gatessucks precisely said it, why not letting him/her answer and take the credits for what he/she said? :) – Öskå Aug 5 '14 at 9:51
@Öskå Fair enough; if / when he posts an answer I'll delete this. In the meantime, I've change it to a community wiki. – Teake Nutma Aug 5 '14 at 9:54
(1/3 x + I y) is different than 1/(3 x + I y). – Artes Aug 5 '14 at 9:55
@Artes That's the OP's fault :P – Öskå Aug 5 '14 at 9:55
@TeakeNutma I don't mean to blame you for this, there are a few questions already answered in the comments which get answered by someone else, I (is it just me?) just think that it's better to ask the original answerer before posting :) – Öskå Aug 5 '14 at 9:57

I believe that you have to split the derivative into two parts (real + imaginary) and then make the corresponding contour plots like this

z[x_, y_] := Abs[((1/3 x + I y) - 2 x)/(I y + 1/x)];
dx[x_, y_] := D[z[x, y], x]
real = Re[ComplexExpand[dx[x, y]]];
img = Im[ComplexExpand[dx[x, y]]];
ContourPlot[real, {x, -2, 2}, {y, -2, 2}, PlotPoints -> 50]
ContourPlot[img, {x, -2, 2}, {y, -2, 2}, PlotPoints -> 50]

The contour of the real part

enter image description here

and that of the imaginary

enter image description here

share|improve this answer
AddSomePictures.png. – Öskå Aug 5 '14 at 9:38
Do I still need to do ContourPlot[{(real)^2 + (img)^2},{x,-2,2},{y,-2,2}] because by defining $z$ to be the absolute function,thus $z$ is real, and so $dx$ must be real? – user44840 Aug 5 '14 at 9:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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