Question

I want to use something like:

ContourPlot [Abs[z-1] == 2]

and to define z as being = x + iy

I saw somewhere an example like that with the With function, but I can’t find it anymore and all my tries are not being successful.

Answer 1

Yes, you can. You just need to be sure to either put the With outside of the ContourPlot:

With[{z = x + I y},
   ContourPlot[Abs[z - 1] == 2, {x, -2, 2}, {y, -2, 2}]]

You can also use Evaluate:

ContourPlot[
  With[{z = x + I y}, Abs[z - 1] == 2] // Evaluate, {x, -2, 2}, {y, -2, 2}]

This is generally the case with plotting functions, which almost always evaluate their arguments in a non-standard way.