Prove this Möbius function maps unit disc to itself bijectively.

Question

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 3.9

I got $(a)$ and $(b)$. My attempt for $(c)$:

First, I interpret that $(c)$ is equivalent to $f_a(D[0,1]), f_{-a}(D[0,1]) \subseteq D[0,1]$.

Next, for $f_a(D[0,1])$, my approach is to let $z=x+iy \in D[0,1]$, i.e. $|z|^2 = x^2+y^2 < 1$ and then plug it in $f_a$:

$$f_a(x+iy) = \frac{(x+iy)-(Re(a)+i Im(a))}{1-(Re(a)+i Im(a))(x+iy)} = \frac{AC+BD}{C^2+D^2} + i\frac{BC-AD}{C^2+D^2}$$

where

$$A := x - Re(a)$$ $$B := y - Im(a)$$ $$C := 1- (xRe(a)+yIm(a))$$ $$D := xIm(a)-yRe(a)$$

Now $$f_a(x+iy) \in D[0,1] \iff |f_a(x+iy)| < 1 \iff |f_a(x+iy)|^2 = \frac{A^2+B^2}{C^2+D^2} < 1$$

$$\iff 0 < (1-|a|^2)(1-|z|^2).$$

Finally, for $f_{-a}(D[0,1])$, I hope that we will similarly have that

$$f_{-a}(x+iy) = \frac{(x+iy)+(Re(a)+i Im(a))}{1+(Re(a)-i Im(a))(x+iy)} \in D[0,1]$$

1. Where have I gone wrong, and why?
2. How could I have more efficiently shown that $f_a(x+iy), f_{-a}(x+iy) \in D[0,1]$? Perhaps polar? Or is this exercise indeed meant to be gory?

Answer 1

What you are supposed to prove is that $f_a$ is a bijection from the open unit disk $\mathbb D$ onto itself. The first thing to check is whether $f_a(\mathbb{D})\subset\mathbb D$. This is true, because, if $|z|<1$, then\begin{align}\left|\frac{z-a}{1-\overline az}\right|^2&=\frac{(z-a)\overline{(z-a)}}{\left(1-\overline az\right)\overline{\left(1-\overline az\right)}}\\&=\frac{|z|^2-\overline az-a\overline z+|a|^2}{1-\overline az-a\overline z+|a|^2|z|^2}\\&=\frac{|z|^2-2\operatorname{Re}\left(\overline az\right)+|a|^2}{1-2\operatorname{Re}\left(\overline az\right)+|a|^2|z|^2}\end{align}and therefore\begin{align}\left|\frac{z-a}{1-\overline az}\right|^2<1&\iff|z|^2-2\operatorname{Re}\left(\overline az\right)+|a|^2<1-2\operatorname{Re}\left(\overline az\right)+|a|^2|z|^2\\&\iff|z|^2+|a|^2<1+|a|^2|z|^2\\&\iff\bigl(1-|z|^2\bigr)\bigl(1-|a|^2\bigr)>0,\end{align}which is true.

So, $f_a$ is indeed a map from $\mathbb D$ into itself. But you have already checked that$$f_a\circ f_{-a}=\operatorname{Id}_{\mathbb D}=f_{-a}\circ f_a.$$The first of these equalities implies that $f_a$ is surjective, whereas the second ont implies that it is injective. So, $f_a$ is a bijection.

Answer 2

I would prove that $|f_a(z)|<1 \Rightarrow |z|<1$.

Assume $|f_a(x)|<1$. Since $$ |f_a(z)| = \left| \frac{z-a}{1-\bar az} \right| = \frac{|z-a|}{|1-\bar az|} = \frac{|z-a|}{|\bar a|\cdot|1/\bar a-z|} $$ we have $$ |z-a| < |a| \cdot |z-1/\bar a| $$ This is now a geometric inequality about lengths in the complex plane. Furthermore the points $a$ and $1/\bar a$ lie on the same ray from the origin, so it makes sense to declare that ray to be the $x$-axis of a new $xy$-coordinate system that we will calculate the lengths in! Then $a$ has coordinates $(A,0)$ for some $A\in(0,1)$, and $1/\bar a$ is $(\frac 1A, 0)$.

Our assumption is now $$ \sqrt{(x-A)^2+y^2} < A \sqrt{(x-\tfrac1A)^2+y^2}. $$ Square both sides and rearrange, and we get $$ (1-A^2)x^2 + (1-A^2)y^2 < 1-A^2 $$ so $x^2+y^2<1$, which is to say $|z|<1$, as desired.

Answer 3

First note that $f_a$ is defined for all points in $\overline D[0,1]$. We will show that $|z|<1\to|f_a(z)|<1$, equivalently;$$|z-a|<|1-\bar az|$$$$|z-a|^2<|1-\bar az|^2$$ $$(z-a)(\bar z-\bar a)<(1-\bar az)(1-a\bar z)$$ $$z\bar z-z\bar a-a\bar z+a\bar a<1-a\bar z-\bar az + \bar aza\bar z$$ $$|z|^2+|a|^2<1+|z|^2|a|^2$$$$|a|^2(1-|z|^2)<1-|z|^2$$$$|a|^2<1,$$ where in the last step, we can divide by $1-|z|^2$ because $|z|<1$. Thus we can consider the restriction $f_a^\star:D[0,1]\to D[0,1]$ of $f_a$. $f_a^\star$ is one-to-one because $f_a$ is one-to-one. Since the inverse $f_{-a}$ of $f_a$ also maps points of $D[0,1]$ to $D[0,1]$, $f_a^\star$ is onto. Thus $f_a^\star$ is a bijection, and $f_a$ maps the unit disc to itself bijectively.