Can Mobius Transformations Map Circles Onto Circles?

  • Thread starter Thread starter sara_87
  • Start date Start date
  • Tags Tags
    Circle Mapping
Click For Summary

Homework Help Overview

The discussion revolves around Mobius transformations and their ability to map circles onto circles. The original poster seeks a general formula for a Mobius transformation that can achieve this mapping, given specific points on the circles involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the properties of Mobius transformations, particularly how they can map circles to lines and vice versa. There is an inquiry into finding a general expression for the transformation that maps a circle to another circle. Questions arise regarding the composition of transformations and the implications of finding an inverse.

Discussion Status

The discussion is active, with participants sharing insights about the inverse of the transformation and its role in achieving the desired mapping. Some guidance has been provided regarding the composition of functions, but there is still uncertainty about the overall approach and how to combine the transformations effectively.

Contextual Notes

Participants are working under the constraints of specific points on the circles and the properties of Mobius transformations. There is an ongoing exploration of definitions and assumptions related to the transformations involved.

sara_87
Messages
748
Reaction score
0

Homework Statement



I know that the Mobius transformation:

g(z) = \frac{z-z_1}{z-z_3}\frac{z_2-z_3}{z_2-z_1}

maps a circle (with points z_1, z_2, z_3 somewhere on the circumference) onto a line.

But, i want a general formula for f(z) that maps a circle (z_1,z_2,z_3) ontp a circle (w_1,w_2,w_3)

Homework Equations





The Attempt at a Solution



Does anyone have any ideas?

Thank you in advance
 
Physics news on Phys.org
The inverse of g(z) maps a line into a circle; find the general expression for its inverse, select three points in a convinient line and compose the two.
 
I found the inverse. It's z= a function in terms of w and z_1, z_2, z_3.

Why does this help? and what do you mean 'compose the two' ?
 
g(z) takes the points (z1,z2,z3) of the circle to (g(z1),g(z2),g(z3)) in a line. Then choose the inverse f(z) such that it takes the points (g(z1),g(z2),g(z3)) in the line, to (w1,w2,w3) in the circle. The composition f(g(z)) will take (z1,z2,z3) in (w1,w2,w3).
 
the inverse f(z) such that it takes the points (g(z1),g(z2),g(z3)) in the line, to (w1,w2,w3) in the circle is:

f(g(z))=\frac{w-w_1}{w-w_3}\frac{w_2-w_3}{w_2-w_1}

so now i have to combine this with the g(z) is gave in the first post?
Why?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K