Mobius Transformations, quick question concepts.

Click For Summary

Homework Help Overview

The discussion revolves around Mobius transformations, specifically the condition ad-bc≠0, which is crucial for the transformation to be well-defined. Participants are exploring the implications of this condition and its relationship to the identity mapping.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand why the condition ad-bc≠0 is necessary and questions whether this could be described by the identity mapping z=f(z). Other participants provide insights into the implications of ad-bc=0, suggesting that it leads to a constant function rather than an identity mapping.

Discussion Status

The discussion is active, with participants questioning the definitions and implications of the Mobius transformation condition. Some guidance has been offered regarding the nature of the transformation when ad-bc=0, but there is no explicit consensus on the interpretation of the identity mapping in this context.

Contextual Notes

Participants are grappling with the definitions and implications of the Mobius transformation, particularly in relation to the identity mapping and the significance of the condition ad-bc≠0. The original poster expresses confusion about the terminology used in textbooks.

binbagsss
Messages
1,291
Reaction score
12
So a mobius transformation is defined as \frac{az+b}{cz+d}=f(z).
Where ad-bc≠0.


My question is just deriving this condition ad-bc≠0.

I understand that the condition describes the case were the mapping leaves all points unchanged. This is described as undefined in some textbooks...(why is this, isn't it then just an identity map?)

But not by setting z=f(z).

I can't see why this condition would not equally be desribed by z=f(z)?

Many Thanks for any assistance.
 
Physics news on Phys.org
If ad-bc=0 then f(z) = b/d which is a constant.
Therefore, the equation y=f(z) can be inverted with respect to z only if ad-bc is not zero.
 
But I can't see why this condition would not equally be desribed by z=f(z)?
 
binbagsss said:
But I can't see why this condition would not equally be desribed by z=f(z)?

I guess I don't see why anyone would describe that mapping where ad-bc=0 as 'leaving all points unchanged'. It doesn't. It maps everything into a constant.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
Replies
4
Views
4K
Replies
10
Views
8K
Replies
4
Views
2K
Replies
2
Views
2K