- #1

docnet

Gold Member

- 799

- 484

- Homework Statement
- .

- Relevant Equations
- .

I just spent 15 minutes to re-type all the Latex again because I lost everything while editing. why does this happen?? This is a huge waste of time.

##f## is entire so ##f## is holomorphic on ##\mathbb{D}∪C##. Also, ##\mathbb{D}∪C## is a connected set. By the maximum principle, ##f## restricted to ##\mathbb{D}∪C## attains its maximum on ##C##. With this, ##|f|≤1## on ##C## leads to ##|f|≤1## in ##\mathbb{D}∪C##

We consider ##\frac{z−a}{−\bar{a}z+1}## The corresponding matrix is $$\begin{bmatrix}1&−a\\−\bar{a}&1\end{bmatrix}$$ The determinant of the matrix is $$1−|a|≠0$$ nonzero because ##|a|<1##. Which means ##\frac{z−a}{−\bar{a}z+1}## is a Möbius transformation.

We decompose the transfomation ##\frac{z−a}{−\bar{a}z+1}##

into the compositions of simple transformations ##f_4(f_3(f_2(f_1(z))))=−\frac{1}{\bar{a}}+\frac{e}{z-\frac{1}{\bar{a}}},e=\frac{|a|−1}{\bar{a}^2}## where

##f_1=z−\frac{1}{\bar{a}}## or translation by ##-\frac{1}{\bar{a}}##

##f_2=\frac{1}{z}## or inversion and reflection with respect to the real axis

##f_3=ez=\frac{|a|−1}{\bar{a}^2}z## or homothety and rotation

##f_4=z-\frac{1}{\bar{a}}## or translation by ##-\frac{1}{\bar{a}}##

questions for everyone:

1. is this is a correct way to prove the original problem?

2. Is the composition supposed to take the unit disk into itself? if so, how do we prove this, and how do we use this to show that ##|\frac{z−a}{−\bar{a}z+1}|\geq 1##? (if this is a correct way to prove the question)

3. It isn't immedietly obvious that it should map the unit disc onto itself, but if so, ##f_3## must un-do the inversion done by ##f_2##, and the translations by ##f_1## and ##f_4## must cancel out. does anyone have any smart explanations for the transformations in common everyday lingo? thank you.

Last edited: