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Re: combining contourplot and regionplot [Addendum]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg98450] Re: [mg98427] combining contourplot and regionplot [Addendum]
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Fri, 10 Apr 2009 04:52:24 -0400 (EDT)
  • Reply-to: hanlonr at cox.net

One other variant

Show[RegionPlot[Abs[Gamma[x + I*y]] > 1, {x, -10, 10}, {y, -10, 10}], 
 ContourPlot[Arg[Gamma[x + I*y]] == 0, {x, -10, 10}, {y, -10, 10}],
 ContourPlot[Arg[Gamma[x + I*y]] == 0, {x, -10, 10}, {y, -10, 10},
  RegionFunction -> Function[{x, y}, Abs[Gamma[x + I*y]] > 1],
  ContourStyle -> Red]]


Bob Hanlon

---- Bob Hanlon <hanlonr at cox.net> wrote: 

=============
Show[RegionPlot[Abs[Gamma[x + I*y]] > 1, {x, -10, 10}, {y, -10, 10}],
 ContourPlot[Arg[Gamma[x + I*y]] == 0, {x, -10, 10}, {y, -10, 10}]]

ContourPlot[Arg[Gamma[x + I*y]] == 0, {x, -10, 10}, {y, -10, 10}, 
 RegionFunction -> Function[{x, y}, Abs[Gamma[x + I*y]] > 1]]


Bob Hanlon

---- Cristina Ballantine <cballant at holycross.edu> wrote: 

=============
Given a complex function (say the Gamma function), I would like to plot 
all points (x,y),  -10<x<10, -10<y<10,  that map to the interval (1, 
infinity). I need to plot all points with Arg[Gamma[x+I*y]]==0 and 
Abs[Gamma[x+I*y]]>1. Solutions to the equation can be plotted with 
ContourPlot. Solutions to the inequality can be plotted with RegionPlot. 
But how do I plot points that satisfy BOTH the equation and the inequality?
 A similar question has been asked on the Forum before but did not receive 
an answer. I am hoping this time someone can help. Thank you.

Cristina



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