Prove this Möbius function maps unit disc to itself bijectively.

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?

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.

• Thanks! I'll analyse later.
– BCLC
Jul 29, 2018 at 1:25

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.

• Thanks. About your first line, this is probably some elementary set theory from the functions chapter in my elementary analysis textbook thing, but What's the difference from the 'finally' part? I meant to do the reverse of the 'next'. I thought such was equivalent to the 'finally'. Anyhoo the reverse of the next would've been equivalent to what you would've done?
– BCLC
Jul 28, 2018 at 17:22
• @BCLC: Since you have proved in part (b) that the inverse function has the same shape, you only need to do one of the directions. Then you can say, "the other direction follows from applying the same argument to $f_a^{-1}=f_{\bar a}$". My point in the first line is just that this direction seems to be easiest to do directly. Jul 28, 2018 at 17:44
• Thanks! I'll analyse later.
– BCLC
Jul 29, 2018 at 1:25

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.

• Thanks! I'll analyse later. On a cursory look, I think I would lean towards accepting this answer.
– BCLC
Jul 29, 2018 at 12:01