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I am using the function ComplexContourPlot to try to get the image of a vertical line under 1/z however I am getting an error.

In this case I would like to see the image of 1/z for the vertical line that goes from 3-3i to 3+3i

ComplexContourPlot[1/z, {z, 3 - 3 I, 3 + 3 I}]

This is the result I am getting:

ComplexContourPlot::plld: Corners for z in {z,3-3 I,3+3 I} must have distinct machine-precision real and imaginary parts.

I have managed to show correctly the circle using a ParametricPlot and working directly with the components u and v.

ParametricPlot[{{3/(3^2 + y^2), -( y/(3^2 + y^2))}}, {y, -20, 20}, 
 PlotLegends -> "Expressions"]

enter image description here

But I want to work directly with Z not divide into the separate real and imaginary parts, how can I do this using Mathematica 12?.

Thank you very much.

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  • $\begingroup$ What do you mean by vertical line under $1/z$? This is a vertical line: ComplexContourPlot[Re[z] == 1, {z, 3}] And for the circle: ComplexContourPlot[Abs[z - 1] == 2, {z, 3}] If you want to visualize mappings, look at this question $\endgroup$
    – Domen
    Sep 3, 2021 at 11:10
  • $\begingroup$ I noticed an error in my original question there was a mistake in how I entered the parameters into ComplexContourPlot. Yes I saw the question, but in that one, is separating real and imaginary parts. I would like to know how to do it directly with z. $\endgroup$
    – Eduardo
    Sep 3, 2021 at 11:16
  • $\begingroup$ Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) 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, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. $\endgroup$
    – bbgodfrey
    Sep 3, 2021 at 14:24

2 Answers 2

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Do you mean the level set of Re[1/z]? Maybe the curve1 is what you want to draw.

curve1 = ComplexContourPlot[Re[1/z] == 3, {z, -3 - 3 I, 3 + 3 I}, 
  PlotPoints -> 80, ContourStyle -> Red]
curve2 = ParametricPlot[{{3/(3^2 + y^2), -(y/(3^2 + y^2))}}, {y, -20, 
   20}, PlotLegends -> "Expressions"]
Show[curve1, curve2, PlotRange -> All]

enter image description here

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ComplexContourPlot draws contour lines of a real function of complex arguments. However, 1/z is not a real function of complex arguments..

To draw the complex values of 1/z over the line 3-3 I to 3+3I you may use AbsArgPlot This will draw the absolute value and color the line according to the argument:

AbsArgPlot[1/(3 + I y), {y, -3, 3}]

enter image description here

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  • $\begingroup$ I thought that in Mathematica every number is a complex number, so I thought by default z would be taken as complex, in ComplexContourPlot. $\endgroup$
    – Eduardo
    Sep 4, 2021 at 21:37

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