1
$\begingroup$

I'm trying to use ParametricPlot to visualize w=1/z. I used:

ParametricPlot[ReIm[1/(x + I y)], {x, -2, 2}, {y, -2, 2}, Mesh -> 20, 
 AxesLabel -> {u, v}, 
 PlotLabel -> "w = \!\(\*FractionBox[\(1\), \(z\)]\)", Frame -> False,
  PlotPoints -> 100, PlotStyle -> Yellow, Exclusions -> None]

but the central region has no mesh (I know that there's a pole in z=0 so I tried Exclusions->None). How can I force ParametricPlot to draw inside the void central region?

enter image description here

Improve this question
$\endgroup$
1
  • $\begingroup$ Using ParametricPlotyou only visualize the parameter plane x,y. Try Plot3D instead. $\endgroup$
    – Ulrich Neumann
    Jul 27, 2021 at 11:49

1 Answer 1

Reset to default
2
$\begingroup$

Since the mapping $w=1/z$ map the circle $|z|=r$ to the circle $|w|=1/r$, so if we want to full fill the $w$ region, we need to set $r$ to infinity,so it is impossible.

Here we set the definition region to Disk[{0, 0}, r] and vary r to demonstrate this.

Table[ParametricPlot[
  ReIm[1/(x + I y)], {x, y} ∈ Disk[{0, 0}, r], 
  MeshFunctions -> {#3 &, #4 &}, 
  Mesh -> {Range[-1, 1, .1], Range[-1, 1, .1]}, BoundaryStyle -> Red, 
  PlotRange -> 6, AxesLabel -> {u, v}, 
  PlotLabel -> "w = \!\(\*FractionBox[\(1\), \(z\)]\)", 
  Frame -> False, PlotPoints -> 100, PlotStyle -> Yellow, 
  Exclusions -> None], {r, {1, 2, 4, 8}}]

enter image description here

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Note that it's important that PlotRange remain fixed during this process; otherwise, Mathematica will "helpfully" change the bounds of the plot, with the result that the hole in the middle appears to remain the same size. $\endgroup$
    – Michael Seifert
    Jul 27, 2021 at 13:59
  • $\begingroup$ Thank you very much $\endgroup$
    – CharlesG
    Jul 27, 2021 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.