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## Homework Statement

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##.

## Homework Equations

## 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?