# Thread: Find a Mobius transformation...

1. ## Find a Mobius transformation...

Let D denote the unit disc |z|<1. Let the point A be inside D. Find a Mobius transformation which maps D bijectively onto itself and the point A to 0.

I'm really not sure how to do this. Are you supposed to pick three points in D and map them to where you want them to go, eg. A -> 0, and put them into some formula? I know a Mobius transformation has the form
f(z) = (az+b)/(cz+d),
I just don't know how to find a,b,c,d.

Thank you! x

2. I think it's:

$T(z)=e^{i\theta}\frac{z-z_0}{1-\overline{z_0}z},\quad \theta\in [0,2\pi]$

3. Ah, now how on earth did you get that? Please explain the method! x

4. Originally Posted by charlie
Ah, now how on earth did you get that? Please explain the method! x
It's a standard fractional linear transformation of the unit disc onto the unit disc. I don't know how to derive it though.

5. ## A solutionH

Here is a solution:

start by noticing that in order to have f(z_o)=0, f is of the form
f(z)=(z-zo)/cz+d).

Now, in order to D to be preserved, the image of the unit circle C by the Mobius transform f must be preserved as well. (Note that when C is preserved, you have D that is preserved or is bijectively sent to the set |z|>1. Since A is inside the unit disc, it's the first possibility that prevails).

Hence, for all $\theta$, |e^(i\theta)-z_o|=|ce^(i\theta)+d|.

Compute this to get the system
|c|^2+|d|^2=1+|z_o|^2
conjugate(c).d=-z_o

Put |c|^2=|zo|^2/|d|^2 in the first equation to get |d|=1.

Finally, cz+d=-z.conjugate(z0)/conjugate(d)+d=d(-z.conjugate(zo)/|d|^2+1)=d(1-zconjugate(zo)), and
f(z)=(1/d)*(z-zo)/(1-zconjugate zo), with |d|=1, which is the solution given above.

I'm sure there must be a more elegant proof.
Sorry for my awful english