Re: Conformal Mapping
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- Subject: [mg128541] Re: Conformal Mapping
- From: Murray Eisenberg <murray at math.umass.edu>
- Date: Fri, 2 Nov 2012 00:43:26 -0400 (EDT)
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On Nov 1, 2012, at 3:19 AM, MaxJ <maxjasper at shaw.ca> wrote: > > I need help finding a Mobius transform such that the region: > > |z-i| < sqrt(2) > && > |z+i| < sqrt(2) > > in z-plane be mapped conformally into a unit circle in w-plane. > > Any help is appreciated very much. > I assume you really did mean "and" rather than "or" in describing the region. This seems more like a mathematics question than a Mathematica question. Mathematica can help peripherally. The two circles bounding the region obviously intersect at complex points z = -1 and z = 1. They intersect the imaginary axis at the points found from: ptBelow=z/.First@Solve[{Abs[z-I]==Sqrt[2],Re[z]==0,Im[z]<0},z]; ptAbove=z/.First@Solve[{Abs[z+I]==Sqrt[2],Re[z]==0,Im[z]>0},z]; pts={ptBelow,ptAbove} {I*(1 - Sqrt[2]), I*(-1 + Sqrt[2])} And you may easy plot the region by using David Park's "Presentations" application, which allows you to express things directly in terms of complex numbers: << Presentations` Draw2D[{ Opacity[0.6], ComplexRegionDraw[Abs[z - I] < Sqrt[2] && Abs[z + I] < Sqrt[2], {z, -2 - 2 I, 2 + 2 I}, BoundaryStyle -> Directive[Thick, Dashed]], PointSize[Large], ComplexPoint /@ pts }, Axes -> True] As to the mathematics: the region is "lens-shaped". Consider what the Moebius transformation you seek does to the boundary -- surely maps it onto the unit circle. Consider the inverse of that transformation. Now apply the theorem that the image of a circle under any Moebius transformation is a circle (in the extended complex plane or, equivalently, on the Riemann sphere). --- Murray Eisenberg murray at math.umass.edu Mathematics & Statistics Dept. Lederle Graduate Research Tower phone 413 549-1020 (H) University of Massachusetts 413 545-2838 (W) 710 North Pleasant Street fax 413 545-1801 Amherst, MA 01003-9305
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- Conformal Mapping
- From: MaxJ <maxjasper@shaw.ca>
- Conformal Mapping