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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