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I am having a brain-not-working day...

Embarrassingly, I can't figure out how to plot a circle on the complex plane with origin z_0 and radio r. Do I just pretend I'm in R^2? Not ideal, since I'd also like to be able to create a table of coordinate points from the circle, too.

Sorry to ask such a basic question. I've got flu and I'm trying to keep myself entertained...

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  • $\begingroup$ Please look under Applications for ComplexListPlot. Get well soon. $\endgroup$
    – Syed
    Dec 13, 2022 at 16:22
  • $\begingroup$ I think the simples is to treat the imaginary axes like an real y axis and then use ` r = 1; z0 = {1, 1}; ParametricPlot[z0 + r ({1, 0} Sin[phi] + {0, 1} Cos[phi]), {phi, 0, 2 Pi}] ` $\endgroup$ Dec 13, 2022 at 17:55
  • $\begingroup$ I don't think this is a well-formed question. Are you really asking for convenient ways to map back-and-forth between the complex plane and the "regular" cartesian plane? In one direction you have ReIm (as well as Abs and Arg). In the other you can just apply Complex. $\endgroup$
    – lericr
    Dec 13, 2022 at 19:20

2 Answers 2

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z0 = 1 + I;
r = 2;
plot1 = ParametricPlot[ReIm[z0 + r E^(I \[Theta])], {\[Theta], 0, 2 \[Pi]}]

enter image description here

points = 
 Table[z0 + r E^(I \[Theta]), {\[Theta], 0, 2 \[Pi], \[Pi]/12}]

enter image description here

plot2 = ListPlot[ReIm[points]];
Show[plot1, plot2]

enter image description here

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With[{z0 = 1 + 2 I, r = 3}, 
 ComplexContourPlot[
  Abs[z - z0] == r, {z, z0 - r (1 + I), z0 + r (1 + I)}]]

enter image description here

With[{z0 = 1 + 2 I, r = 3}, 
 ComplexContourPlot[
  Abs[z - z0] == r, {z, z0 - r (1 + I), z0 + r (1 + I)}, 
  MeshFunctions -> {Abs[# - z0] &, Arg[# - z0] &}, Mesh -> {{r}, 20}, 
  MeshStyle -> Directive[Red, AbsolutePointSize[8]]]]

enter image description here

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