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