2
$\begingroup$

This question already has an answer here:

With[{f = # + 1/# &, center = 1/3 + 3 I/2, radius = 4/3},ParametricPlot[Through[{Re,       Im}[f[center + r Exp[I \[Theta]]]]], {r, 0, radius}, {\[Theta], -\[Pi], \[Pi]}, PlotPoints -> 30, PlotRange -> All, MaxRecursion -> 3]]

in Version 9,the output is:

enter image description here

but in Version 10:

enter image description here

Is it a bug in Version 10?

the code links here.

$\endgroup$

marked as duplicate by Michael E2, Dr. belisarius, Bob Hanlon, bobthechemist, Öskå Dec 11 '14 at 22:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

5
$\begingroup$

Use PlotTheme -> "Classic" to get V 9 output:

Mathematica graphics

With[
 {f = # + 1/# &, center = 1/3 + 3 I/2, radius = 4/3},
 ParametricPlot[Through[{Re, Im}[f[center + r Exp[I \[Theta]]]]], {r, 0, radius}, 
  {\[Theta], -\[Pi], \[Pi]}, PlotPoints -> 30, 
  PlotRange -> All, MaxRecursion -> 3, PlotTheme -> "Classic"]
 ]

Mathematica graphics

I am not sure why, might be a bug

$\endgroup$
  • $\begingroup$ Thanks a lot! I don't know PlotTheme->"Classic" before this post. 😇 $\endgroup$ – partida Nov 28 '14 at 10:42
3
$\begingroup$

On 10.0 for Mac OS X x86 (64-bit) (September 10, 2014) one uses:

With[{f = # + 1/# &, center = 1/3 + 3 I/2, radius = 4/3}, 
 ParametricPlot[
  Through[{Re, Im}[f[center + r Exp[I \[Theta]]]]], {r, 0, 
   radius}, {\[Theta], -\[Pi], \[Pi]}, PlotPoints -> 55, 
  PlotRange -> All, Mesh -> Full]]

enter image description here

So, Mesh seems to do the trick.

$\endgroup$
  • 2
    $\begingroup$ Mesh -> Automatic allows to reproduce the output of version 9 exactly (only colors are different). It is despite the fact that by default Options[ParametricPlot, Mesh] returns {Mesh -> Automatic}. $\endgroup$ – Alexey Popkov Nov 28 '14 at 13:56

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