Conformal Mapping

[MaxJ]

Hi folks,

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.

MaxJ

[Murray Eisenberg]

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 mur...@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

[Andrzej Kozlowski]

You most certainly do not need any external packages (you never really need them, don't believe when they tell you otherwise). To get a quick picture of your region just use:


RegionPlot[
Abs[x + I y - I] < Sqrt[2] && Abs[x + I y + I] < Sqrt[2], {x, -2,
2}, {y, -2, 2}]

You can make it much more fancy if you want, but for most purposes this is enough. It's both quicker and much cheaper than the suggested method.

Murray's mathematical solution is, of course, fine although he did not give you the explicit mapping. Actually, you can automate almost the entire process by using the cross ratio. First, define the cross ratio:

crossRatio[z_, q_, r_, s_] := (z - q) (r - s)/((z - s) (r - q))

Next, we take three points on the boundary of the lens and map them onto three points on the boundary of the unit circle. Since Murray has already shown how to find the first three points, we can take:

points1 = {-1, 1, (-1 + Sqrt[2]) I};

For the three points on the unit circle we can take

points2 = {I, -I, 1};

The order is important. Also, since the Moebius transformation does not map the lens onto the unit disk (only into), you can find more than one mapping that wil accomplish the task. Anyway, here is how we find a mapping:

ff[z_] = w /.
Solve[crossRatio[z, Sequence @@ points1] ==
crossRatio[w, Sequence @@ points2], w][[1]] // Simplify;

Lets check if our ff does really what we want, in other words, maps points1 onto points2:

ff/@points1//Simplify
{I,-I,1}

So we have a Moebius transformation (or a linear fractional transformation) which maps three points on the boundary of the lens onto three points on the unit circle. It is easy to check that the interior of the lens goes into the interior of the circle. To see the image of the interior we need to find the inverse Moebius transform:

gg[w_]=z/.Solve[ff[z]==w,z][[1]];

Check again that this works fine on the boundary points:

gg/@points1//Simplify
{-1,1,I (Sqrt[2]-1)}

We can now again use RegionPlot to see the image of the lens:

RegionPlot[
Abs[gg[x + I y ] - I] < Sqrt[2] &&
Abs[gg[x + I y ] + I] < Sqrt[2], {x, -2, 2}, {y, -2, 2}]

You can also check that the image of the half-disk under the inverse Moebius transform gg is indeed your lens:

RegionPlot[
Abs[ff[x + I y]] < 1 && Re[ff[x + I y]] > 0, {x, -2, 2}, {y, -2, 2}]


Andrzej Kozlowski

[Murray Eisenberg]

On Nov 4, 2012, at 12:44 AM, Andrzej Kozlowski <akozl...@gmail.com> wrote:

>
> On 2 Nov 2012, at 05:43, Murray Eisenberg <mur...@math.umass.edu> wrote:
>
>>
>> On Nov 1, 2012, at 3:19 AM, MaxJ <maxj...@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.
>>
>>

>> I assume you really did mean "and" rather than "or" in describing the region.
>>

>> 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).
>>
>

Yes, of course one can map the lens-shaped region _into_ the unit circle, as you showed. I misread "into" as "onto".

As to drawing the region: Yes, of course one can do it with out-of-the-box Mathematica. But it seems counterintuitive to have to plot a figure involving a complex-valued function of a complex variable by breaking complex numbers z apart into their real and imaginary parts x and y. After all, for calculations Mathematica "wants" numbers to be complex rather than real! What Park's "Presentations" allows is to work directly in complex terms for plotting. the "Presentations" primitive ComplexRegionDraw is just the tip of the iceberg in complex facilities provided.

[Andrzej Kozlowski]

On 4 Nov 2012, at 16:29, Murray Eisenberg <mur...@math.umass.edu> wrote:
>
> As to drawing the region: Yes, of course one can do it with
out-of-the-box Mathematica. But it seems counterintuitive to have to
plot a figure involving a complex-valued function of a complex variable
by breaking complex numbers z apart into their real and imaginary parts
x and y. After all, for calculations Mathematica "wants" numbers to be
complex rather than real! What Park's "Presentations" allows is to work
directly in complex terms for plotting. the "Presentations" primitive
ComplexRegionDraw is just the tip of the iceberg in complex facilities
provided.
>

Maybe, but every Mathematica user ought to acquire enough basic skill to
overcome this supposed "counter-intuitiveness". After all, it is hardly
honest to encourage people to use Mathematica by telling them how
powerful it is and how much simpler than, say, learning C, and then the
moment they try to solve a simple mathematical problem tell them that
the best thing to do is to buy an add-on package because Mathematica
itself is what =85 too complex for them o learn?

And while you are recommending them to get this package you omit to
mention that they are not going to be able to share the code they
produced with its help with anyone who does not have the package, or
embed it in a CDF, etc. Furthermore, by relying on such a package are
making themselves dependent on it's author who one day may not want to
or more likely be able to make it compatible with future versions of
Mathematica. I would think that these are sufficient reasons to hesitate
before recommending it to anyone but people who really need it and have
no other alternative and this case I certainly do not see as belonging
to this category.

Andrzej Kozlowski

[Murray Eisenberg]

That plotting a complex region or function with bare Mathematica requires resort to real and complex parts just shows an unfortunate shortcoming of Mathematica. And a kind of inconsistency, since by default in algebraic calculations numbers are regarded as complex.

[Roland Franzius]

The boundaries are two circles with centers at +-i and radius^2 = 2.
Consequently the contours are passing through the six points on the
squares with vertices

+-1, 2i +-1 and +-1 -2i +-1

Consequently the lens shaped area in question is bounded by two
symmetric quarter circles with center at +-i intersecting at +-1 with an
angle of pi/2.

We conclude that the identity w=z is a conformal map of the lens into
the inner of the unit circle.

The construction of a conformal map onto the unit circle, transforming
to straight lines the pi/2 vertices at +-1, needs square roots centered
at the vertices +-1.

This task could generate homework for up to one day approximately.

--

Roland Franzius

[Andrzej Kozlowski]

On 6 Nov 2012, at 00:39, Roland Franzius <roland....@uos.de> wrote:

> Am 01.11.2012 08:22, schrieb MaxJ:

> The boundaries are two circles with centers at +-i and radius^2 = 2.
> Consequently the contours are passing through the six points on the
> squares with vertices
>
> +-1, 2i +-1 and +-1 -2i +-1
>
> Consequently the lens shaped area in question is bounded by two
> symmetric quarter circles with center at +-i intersecting at +-1 with an
> angle of pi/2.
>
> We conclude that the identity w=z is a conformal map of the lens into
> the inner of the unit circle.
>
> The construction of a conformal map onto the unit circle, transforming

> to straight lines the pi/2 vertices at +-1, needs square roots centered
> at the vertices +-1.
>
> This task could generate homework for up to one day approximately.
>
> --
>
> Roland Franzius
>

There is no need for square roots centered at vertices; ordinary squaring will suffice. And it's less then one day's homework; without Mathematica a maximum of 15 minutes I think, with Mathematica a lot less.

The standard way to deal with a lens is to use a Moebius transform to send one vertex to 0 and the other to infinity. In this case the map =09

f[z_]:=(z+1)/(z-1)

will do it. We get the angle"

g = InverseFunction[f];

RegionPlot[
Abs[g[x + I y] - I] < Sqrt[2] &&
Abs[g[x + I y ] + I] < Sqrt[2], {x, -2, 2}, {y, -2, 2},
Frame -> False, Axes -> True]

Next, we rotate it to the right by -3Pi I/4 and use squaring (not square root). We will get the upper half plane. We can now map the upper-half plane to the unit disk by a Moebius transformation - fro example

kk[z_]:=(z-I)/(z+I)

The final map is:

gg[z_]=Simplify[kk[(f[z]Exp[-3Pi/4 I])^2]]
(2 z)/(z^2+1)

We can check this graphically:

RegionPlot[Abs[gg[x + I y]] < 1, {x, -2, 2}, {y, -2, 2}]

We see the lens plus some additional region. Both the lens and the additional region are mapped onto the unit disk. In order to construct the inverse map from the unit disk to the lens we need, of course, a square root.

Andrzej Kozlowski

[Roland Franzius]

Homework and time was meant for students. In this group, to find the
solution needed about one week.

Thats of course perfect, the students will now have learnt about the
solution after that week by their instructor, probably.

> The standard way to deal with a lens is to use a Moebius transform to send one vertex to 0 and the other to infinity. In this case the map =09
>
> f[z_]:=(z+1)/(z-1)
>
> will do it. We get the angle"
>
> g = InverseFunction[f];
>
> RegionPlot[
> Abs[g[x + I y] - I]< Sqrt[2]&&
> Abs[g[x + I y ] + I]< Sqrt[2], {x, -2, 2}, {y, -2, 2},
> Frame -> False, Axes -> True]
>
> Next, we rotate it to the right by -3Pi I/4 and use squaring (not square root). We will get the upper half plane. We can now map the upper-half plane to the unit disk by a Moebius transformation - fro example
>
> kk[z_]:=(z-I)/(z+I)
>
> The final map is:
>
> gg[z_]=Simplify[kk[(f[z]Exp[-3Pi/4 I])^2]]
> (2 z)/(z^2+1)
>
> We can check this graphically:
>
> RegionPlot[Abs[gg[x + I y]]< 1, {x, -2, 2}, {y, -2, 2}]
>
> We see the lens plus some additional region. Both the lens and the additional region are mapped onto the unit disk. In order to construct the inverse map from the unit disk to the lens we need, of course, a square root.

Yes, approved;-)

--

Roland Franzius

[djmpark]

In response to a series of postings on MathGroup that derived from a simple
conformal mapping question
(http://forums.wolfram.com/mathgroup/archive/2012/Nov/msg00004.htm,
http://forums.wolfram.com/mathgroup/archive/2012/Nov/msg00011.html,

http://forums.wolfram.com/mathgroup/archive/2012/Nov/msg00034.html )

Murray Eisenberg and I have produced a short didactic notebook showing a
simple specific solution of the Laplace equation on a lens-shaped domain.
The notebook and PDF have been posted at Peter Lindsay's site at the
Mathematics Department of St Andrews University.



http://www.st-andrews.ac.uk/~pl10/c/djmpark/



This was done primarily to review and better understand the steps in this
process and also to illustrate how the complex graphics routines of the
Presentations application make it relatively easy to produce high quality
graphics directly in terms of complex functions and variables. Of course all
of the graphics in the notebook could be made in regular Mathematica. It is
simply a matter of what value one puts on one's time, whether one wants to
think more about mathematics or about programming, and on whether one would,
in fact, make the graphics.





David Park

djm...@comcast.net

http://home.comcast.net/~djmpark/index.html

[MaxJ]

Thanks a lot to all you folks for your superb & enlightening comments and solutions.

QUESTION:

Based on your solutions, shall I conclude that the lens region can only be mapped into half of unit disk?

Thanks.

Max.

[Andrzej Kozlowski]

It depends on what sort of mapping you are asking for. The lens can be
mapped conformally onto the unit disk, of course. In fact, by the
Riemann mapping theorem any open simply connected region (except the
entire complex plane) can be mapped conformally onto the unit disk.
However, such a mapping cannot be a Moebius transformation (which is the
same as a "linear fractional transformation" and which is what you
explicitly asked for in your question). A Moebius transformation mapps
(generalized) circles to circles and hence it clearly cannot map a lens
onto a disk. But of course you can map a lens into a unit disk be means
of a Moebius transformation in lots of different ways. In fact, since a
Moebius transformation is completely determined by choosing three
distinct points in the source space and three distinct points as their
images in the target space, the method I posted makes it possible to
find all ways of mapping a lens into the unit disk by a Moebius
transformation. I chose the Moebius map that sends the lens onto a half
disk only because it seemed the most "attractive" of an infinitely many
"into" Moebius mappings.
Of course none of these mappings is onto. The the biholomorphic mapping
which I constructed in another post is a rational map of degree 2 -
hence not a Moebius map.

Andrzej Kozlowski