# How do I plot the differential of this function?

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.

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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

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)];

z[x,y]

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}]


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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

and that of the imaginary

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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