- 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}]

- 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]

- 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]

- 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}]]

- 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]]

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]]

ContourPlotis 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 toSequence[2 z, z -> Iy + x, Abs[z] == 1]Also note thatIyis a variable not a product, you needI yorI*y– Bob Hanlon Jan 21 '22 at 14:13