There is a simple procedure for determining fractional linear transformations which map a given set of three points to another set of three points. The problem is as follows: There are three distinct points in the complex plane, z_{1},z_{2}, and z_{3} and it is desired to find a fractional linear transformation such that z_{i} → w_{i} for i = 1,2,3 where here w_{1},w_{2}, and w_{3} are three distinct points in the complex plane. Then the procedure says that to find the desired fractional linear transformation solve the following equation for w.

The result will be a fractional linear transformation with the desired properties.
Why should this procedure work? Here is a heuristic argument to indicate why you would expect this to happen rather than a rigorous proof. The reader may want to tighten the argument to give a proof. First suppose z = z_{1}. Then the right side equals zero and so the left side also must equal zero. However, this requires w = w_{1}. Next suppose z = z_{2}. Then the right side equals 1. To get a 1 on the left, you need w = w_{2}. Finally suppose z = z_{3}. Then the right side involves division by 0. To get the same bad behavior, on the left, you need w = w_{3}.
Example 17.5.1 Let z_{1} = 0,z_{2} = 1, and z_{3} = 2 and let w_{1} = 0,w_{2} = i, and w_{3} = 2i.
Then the equation to solve is

Solving this yields w = iz which clearly works.
Example 17.5.2 Let ξ ∈ ℂ and suppose Im

Let U ≡ ℂ ∖

for z. This yields

As long as w≠1, this gives a solution because z≠. (

and so you would need to have
Thus this is an example of an analytic function defined on a connected open set, in this case, the upper half plane such that the image of this analytic function is the unit ball. This begs the question of which connected open sets can be mapped one to one by an analytic function onto the unit ball. It turns out that every simply connected open set will have this property. Simply connected means that if you have any simple closed curve contained in the region, then the inside of the simple closed curve is in the region.
Proof: First of all, for

after a few computations. If I show that ϕ_{α} maps B
Consider

where the first equality is obtained by multiplying by

whenever
It only remains to verify the assertions about the derivatives. Long division gives ϕ_{α}
The next lemma is called the Schwarz lemma. It was presented earlier in the exercises.
 (17.4) 
and
 (17.5) 
If equality holds in 15.36 then there exists λ ∈ ℂ with
 (17.6) 
Proof: First note that by assumption, F

Then letting r → 1,

this shows 17.4 and it also verifies 17.5 on taking the limit as z → 0. If equality holds in 17.5, then
Rudin [109] gives a memorable description of what this lemma says. It says that if an analytic function maps the unit ball to itself, keeping 0 fixed, then it must do one of two things, either be a rotation or move all points closer to 0. (This second part follows in case
In the next section, the problem of considering which regions can be mapped onto the unit ball by a one to one analytic function will be considered. Some can and some can’t. It turns out that the main result in this subject is the Riemann mapping theorem which says that any simply connected connected open set can be so mapped onto the unit ball. A key result in showing this is the fact that such regions have something called the square root property.
Definition 17.5.5 A region, Ω has the square root property if whenever f,
The following lemma says that every simply connected region which is not all of ℂ has the square root property.
Proof: Let f and