# Find analytic function that maps conformal

1. Apr 19, 2014

### skrat

1. The problem statement, all variables and given/known data
Find analytic function $f$ that maps $Re(z)>0$ conformal on open unit disk; so that the point $0$ is a map of point:
a) $1$,
b)$3$.

2. Relevant equations

3. The attempt at a solution

I am having some troubles with these möbius transformations... Let me first start with examples and later continue with my solutions of the original problem.

The way I understand it, it is much easier to find a function that maps open unit disk into right half-plane. Therefore I tried to imagine that we at some point rip the unit disk apart; let that be at point $1$, and send it to $\infty$. So $f(1)=\infty$. Now point on the opposite site of $1$, that's $-1$ should by this logic go to $0$ and accordingly $f(i)=i$.

This would than give me $f(z)=\frac{az+b}{cz+d}=\frac{z-i}{-z-i}$. Now I would of course still have to find the inverse function, BUT lets go to problem a):

Following the written above (which obviously has to be wrong - I just don't know why). The problem wants that $f(1)=0$, than by this logic $f(-1)=\infty$ and $f(i)=i$.

However, calculating $a$, $b$, $c$ and $d$ give me $c=d=0$. NOT possible.

Since I don't want anybody to say that I didn't even try to find a solution, here is one that should work but I do not understand:

In my notes it says that for $f(i)=0$, $f(1)=i$ and $f(-i)=\infty$ I get a function $f(z)=\frac{z-1}{-z-i}$. However, this maps $i$ to $0$ instead of $1$.

Therefore I decided to go with $f(z)=f_2\circ f_1$, where $f_1(z)=-1+i$ and $f_2(z)=\frac{z-1}{-z-i}$. Both together give me $f(z)=\frac{z-1}{-z+1-2i}$.

I guess my question here is: how do I determine where to map those first three points?

2. Apr 20, 2014

### haruspex

I don't understand why you want f(i) = i. What you know is |f(iy)|=1.

Last edited: Apr 20, 2014
3. Apr 21, 2014

### skrat

Only because I need three points and because I can't see why that would be impossible. No other reason.
How do I know that |f(iy)|=1?

4. Apr 21, 2014

### haruspex

It is possibly the case, but you don't know that it is, so it might lead to a wrong answer.
I just realised I hadn't read the OP acurately. Your f is actually the inverse function, so I should have written |f-1(iy)|=1.
The original region and the target region are both open regions, so you need to map the boundary of one to the boundary of the other. The boundary of the source region includes the imaginary axis, iy, so this must be mapped to the circle |z|=1.
Also, I think choosing f(1)=∞ may be making life hard.
Are you familiar with polar inversion (I think that's what it's called)? You pick a point O and a radius r, and replace each point X in your diagram with a point X' where O, X and X' are collinear and OX*OX' = r2. This maps circles to circles, including the special cases of circles through O and straight lines (not through O) which get mapped to each other.
To make this relate to analytic functions we can add another twist, reflection. Inversion wrt the origin, radius 1, plus a reflection in the real axis, maps z to 1/z.
I mention this because I found it much easier to solve this problem by thinking terms of an inversion. In particular, I mapped +∞ to 0 first, then just subtracted 1 to get it to -1.
Sorry, this sounds all a bit confusing, but I'm trying to put you on the right track without giving away the whole answer.

5. Apr 21, 2014

### skrat

This method may be much easier, but sadly this is not something we have ever done.
We only used linear transformations (rotations, translations and (something I don't know the english word :D but can be written as $f(z)=az$)) and Möbius transformation.

I managed to find something here: http://en.wikipedia.org/wiki/Möbius_transformation#Specifying_a_transformation_by_three_points but this looks way more complicated than what we did.

Like I said, we usually determined three points and their maps and than calculate a,b,c and d. However I don't really how to determine them, because every time I seem to determine them wrong! -.-
If necessary I have a couple more examples which may be easier to understand than the original problem.

I do apologize for not taking a closer look into your method, however I do believe it would be better for me to master the method we used and than search for easier ones.

6. Apr 21, 2014

### haruspex

Ok, but I think trying to guess a map for i is the stumbling block. Try guessing maps (in the half plane to disk direction) for 0, 1 and ∞.