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I found the solution to plot loci of complex numbers here, e.g. a locus of points of $z$ given $|z|=1$ is a circle centered at the origin with a radius 1.

But I am struggling to plot transformations, e.g. $T: w=2z$ would be an enlargement of the circle by a factor of 2.

I am trying something like this, but not very

ContourPlot[Evaluate[w /. w -> 2*z, z -> x + Iy, Abs[z] == 1],
{x, -4,4},
{y, -4, 4},
Axes -> True]
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    $\begingroup$ The argument to ContourPlot is not what you think. Look at it before to try to plot it. Evaluate[w /. w -> 2*z, z -> x + Iy, Abs[z] == 1] evaluates to Sequence[2 z, z -> Iy + x, Abs[z] == 1] Also note that Iy is a variable not a product, you need I y or I*y $\endgroup$
    – Bob Hanlon
    Jan 21 at 14:13

2 Answers 2

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So if you want to plot points that fulfill the equation |z| = w with w = 2. You can get the required result with

ContourPlot[Evaluate[Abs[z] == w /. { z -> x + I y, w -> 2}], {x, -3, 3}, {y, -3, 3}, Axes -> True]

contour of |z|=2

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    $\begingroup$ {y, 3, 3} should read {y, -3, 3} $\endgroup$
    – Daniel Huber
    Jan 21 at 14:10
  • $\begingroup$ Not really what I'm looking for. I am looking for syntax where $z$ is as defined, so I can play around with $w=f(z)$... i.e. a general solution, not specific to this example. $\endgroup$
    – Mihail
    Jan 21 at 14:20
  • $\begingroup$ You can use any equation in the contour plot you want. If you want to define a function f[z] you could use something like this: f[z_] := 3 (z - 1)^2; w = f[z] /. {z -> x + I y}; ContourPlot[Abs[w] == 1, {x, -3, 3}, {y, -3, 3}, Axes -> True] $\endgroup$
    – Mathias
    Jan 21 at 14:44
  • $\begingroup$ Still not what I am looking for. I need to be able to define $z$ as a condition for the locus instead of an equation. See the link in Q on how it was done for a plot w/o transformation. I need to extend that to a plot w/ transformation and be able to define the transformation separately. $\endgroup$
    – Mihail
    Jan 21 at 15:16
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  • Since w=2z means that z=w/2,so we can plot Abs[w/2]==1
ComplexContourPlot[Abs[w/2] == 1, {w, -4 - 4 I, 4 + 4 I}]

enter image description here

  • Another example by using ComplexContourPlot.

If you want to plot the mapping w=(z+1)^2 under the condition Abs[z]==1,we can solve z=Sqrt[w]-1 and plot Abs[Sqrt[w]-1]==1

ComplexContourPlot[Abs[Sqrt[w] - 1] == 1, {w, -4 - 4 I, 4 + 4 I}, 
 PlotPoints -> 50, MaxRecursion -> 4]

enter image description here

  • The same as ParametricRegion if we add the restricted condition Abs[z]==1.
reg = Block[{z = x + I*y}, 
   ParametricRegion[{ReIm[(z + 1)^2], Abs[z] == 1}, {x, y}]];
Region[reg, Axes -> True, AspectRatio -> Automatic, 
 BaseStyle -> Thick]

enter image description here

  • If we rewrite Abs[z]==1 by z=Exp[I*t],0<=t<=2 π, then ParametricPlot also work.
Block[{z = Exp[I*t]}, 
 ParametricPlot[ReIm[(z + 1)^2], {t, 0, 2 π}]]

enter image description here

  • Set MeshFunctions -> {Norm[{#3, #4}] &}, Mesh -> {{1}} also means that Abs[z]==1.
Block[{z = x + I*y}, 
 ParametricPlot[ReIm[(z + 1)^2], {x, -1, 1}, {y, -1, 1}, 
  MeshFunctions -> {Norm[{#3, #4}] &}, Mesh -> {{1}}, 
  MeshStyle -> {Thick, Red}, PlotStyle -> None, BoundaryStyle -> None,
   PlotPoints -> 50]]

enter image description here

  • Abs[z]==1 is the boundary of Disk[], we can use ParametricPlot again.
Block[{z = x + I*y}, 
 ParametricPlot[ReIm[(z + 1)^2], {x, y} ∈ Disk[], 
  BoundaryStyle -> Red, PlotStyle -> None, PlotPoints -> 50]]

enter image description here

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