How Do Mobius Transformations Map Complex Domains?

  • Thread starter Thread starter sgcbell2
  • Start date Start date
  • Tags Tags
    Transformations
Click For Summary
SUMMARY

The discussion focuses on Mobius Transformations in the context of conformal mappings, specifically finding the transformation that maps the points 1, i, and 0 to ∞, 0, and 1, respectively. The transformation identified is (z-i)/i(z-1). Participants explore how to determine the image of the domain { z : 0 < x < t} under this transformation, with emphasis on evaluating boundary values to visualize the mapped shape. The transformation effectively maps complex numbers, confirming the expected behavior of the function.

PREREQUISITES
  • Understanding of Mobius Transformations
  • Familiarity with complex numbers and their properties
  • Knowledge of conformal mappings
  • Basic skills in evaluating complex functions
NEXT STEPS
  • Research the properties of Mobius Transformations in complex analysis
  • Learn how to visualize complex mappings using software like GeoGebra
  • Explore the concept of conformal mappings and their applications
  • Study the implications of boundary value problems in complex functions
USEFUL FOR

Mathematicians, students of complex analysis, and anyone interested in the applications of Mobius Transformations in conformal mapping techniques.

sgcbell2
Messages
3
Reaction score
0
Hi, I am currently working on problem to do with conformal mappings and Mobius Transformations.

My problem is:
Find the Mobius Transformation which carries the points 1, i, 0 to the points ∞, 0, 1 (in precisely this order). Find the image of the domain { z : 0 < x < t} under this Mobius Transformation, where t > 0 is some fixed positive real number.

I have found this Mobius transformation to be (z-i)/i(z-1).

I thought that in order to find the image of the domain given above, I should input different values along the boundaries x=0 and x=t into this mobius transformations to see where they are mapped to. I have done this but I'm finding it hard to see the shape of the image.

Maybe I am doing something wrong?
 
Physics news on Phys.org
Hello sg and welcome to PF.
I'm not really qualified re conformal mapping, but from what I know from complex numbers, I see the function you found transforms 1+0i to ##\infty##, 0+i to 0 and 0 + 0i to 1+0i.

I see x+0i is mapped to 1/(1-x) + ix/(1-x) which satisfies Re - I am = 1 , so I would expect a line I am = Re - 1
 
  • Like
Likes   Reactions: 1 person
That makes a lot of sense! Thank you
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
4
Views
4K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K