Mobius Transformations and Stereographic Projections

Click For Summary
The discussion focuses on Mobius transformations and their properties, specifically regarding fixed points in the context of stereographic projections. It establishes that any Mobius transformation not equal to 1 on the extended complex plane has one or two fixed points. The transformation corresponding to a rotation of S^2 through a nonzero angle has exactly two fixed points, z_1 and z_2, where z_2 is defined as -1/z_1*. The conversation also touches on proving that if a Mobius transformation has two fixed points satisfying this relationship, it either corresponds to a rotation of S^2 or has an attractive fixed point. The original poster initially sought help but later resolved their queries independently.
Mathmos6
Messages
76
Reaction score
0

Homework Statement



Hi all - I've been battering away at this for an hour or so, and was hoping someone else could lend a hand!

Q: Show that any Mobius transformation T not equal to 1 on \mathbb{C}_{\infinity} has 1 or 2 fixed points. (Done) Show that the Mobius transformation corresponding (under the stereographic projection map) to a rotation of S^2 through a nonzero angle has exactly 2 fixed points z_1 and z_2, where z_2=\frac{-1}{z_1^*}. If now T is a Mobius Transformation with 2 fixed points z_1 and z_2 satisfying z_2=\frac{-1}{z_1^*}, prove that either T corresponds to a rotation of S^2, or one of the fixed points, say z_1, is an attractive fixed point (i.e. for z not equal to z_2, T^nz_1 \to z_1 as n \to \infty).

Now I believe I've shown that the Mobius transformation corresponding to a rotation is mapped to mobius transformations with 2 fixed points, but I'm unsure as to how to show that z_2=\frac{-1}{z_1^*}, and I'm extra extra unsure how to show the later point about attractive fixed points! Please do reply, the more you can help me the better, and I certainly do need it! Many thanks, Mathmos6
 
Physics news on Phys.org
No worries, I got it sorted on my own anyways.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
2K
Replies
26
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K