# Homework Help: Finding Mobieus mapping

1. Sep 27, 2009

### soopo

1. The problem statement, all variables and given/known data

Find the Mobieus mapping that maps { z e C, |z| <= 1 } to a disk {z e C, |z - 1| <= 1} in a real axis.

3. The attempt at a solution

I have had an idea that Mobieus mapping is from C to C such that it is a homeomorfism and it has an inverse mapping.

I am not sure how you can use it for the mapping.

2. Sep 27, 2009

### Dick

I'm not sure you do know what a mobius map is. You should probably look it up. It's a map of the form f(z)=(az+b)/(cz+d) with a, b, c and d complex constants. All you need is a translation.

3. Oct 3, 2009

### soopo

1. to take the corners of the initial disk such as { (-1,0), (0,1), (1,0) } in x -axis
2. to map them to the corner points { (1,0), (2,1), (3,0) } in the x-axis

I can change the Mobieus mapping from
(az + b) / (cz + d) to (z + b) / (cz + d)
by canceling the "stabilising" variable "a".

I should apparently have the z -coordinate in the mappings too.
However, I do have three points which suggests me that I can find the Mobieus map:

(-1, 0, 1) -> (1, 2, 3)

I get the following mapping by calculating the equations
M = z + 2,
when b=2, c=0 and d=1.

4. Oct 4, 2009

### Dick

I don't understand what you are doing. What's wrong with f(z)=z+1? Unless I'm completely wrong about what the question is. You are just mapping one disk to another disk, right? What does 'in a real axis' mean?

5. Oct 4, 2009

### soopo

It seems to mean that I need to map the circle to a real line such that the inner points of the circle of the circle are above the real axis while the corner points are on the axis.

f(z) = z + 1 seems to be too ok.
Your mapping has the different selection of the corner points so you get different values for b, c and d.

I get the mapping by first selecting the following corner points of the circle and then mapping them to (1,2,3) such that (-1, 0, 1) -> (1, 2, 3).

This selection gives me the mobius mapping: f(z) = z + 2.

6. Oct 4, 2009

### Dick

Ok, so you want to map the unit disk to the upper half plane? f(z)=z+2 definitely doesn't work. It generally easier to construct these maps by combining a few basic ones. See what the image of unit disk is under the map g(z)=1/(1-z). It's maps the disk to a half-plane. It's not the half-plane you want, but can you change the mapping so that it is?