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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.

Eduardo
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  • 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 – Domen Sep 03 '21 at 11:10
  • 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. – Eduardo Sep 03 '21 at 11:16
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2 Answers2

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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

cvgmt
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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

Daniel Huber
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  • I thought that in Mathematica every number is a complex number, so I thought by default z would be taken as complex, in ComplexContourPlot. – Eduardo Sep 04 '21 at 21:37