Consider $a \in \mathbb{C}$, where $|a|<1$ and $f_a(z)=\frac {z-a}{1-\overline az}$.
(a) Show that $f_a(z)$ is a Mobius transformation.
(b) Show that $f_a^{-1}(z)=f_{-a}(z)$
(c) Prove that $f_a(z)$ maps the unit disks $D[0,1]$ to itself in a bijective fashion.
Answer:
(a) It can be shown that $f_a(z)$ is a Mobius transformation since $(1)(1)-(a)(\bar a)=1-a\overline a=1-|a|^2>0 \neq 0$.
(b) It can be easily shown that $f_a^{-1}(z)=f_{-a}(z)$. Consider $w=\frac {z-a}{1-\overline az}$, then $z=\frac {w+a}{1+w\overline a}$. So $f_a^{-1}(z)=\frac{z+a}{1+\overline az}$. In addition, $f_{-a}(z)=\frac {z-(-a)}{1-\overline {(-a)}z}=\frac {z+a}{1+\overline az}$. Hence $f_a^{-1}(z)=f_{-a}(z)$.
I have a problem with (c) though, as the solution provided by the lecturer is quite complicated and I feel that they should be an easier way to prove the bijectivity. The solution he provided is as follows:
Consider $|z|=r, 0 \leq r \lt 1$. Then $ \lvert \frac{w+a}{\overline aw+1} \rvert =r$ by using $f_a^{-1}(z)$. So $\lvert w+a \rvert=r\lvert \overline aw+1 \rvert$. By squaring both sides, and using properties of conjugates, $(w+a)(\overline w+\overline a)=r^2(\overline aw+1)(\overline aw+1)$.
Expanding this would give;
$w\overline w+w\overline a+a \overline w+a\overline a=r^2(a\overline aw\overline w+\overline aw+a\overline w+1)$. By shifting the terms, we would obtain $r^2 - a\overline a=w\overline w+w\overline a+w\overline a-r^2a\overline aw\overline w-r^2\overline aw-r^2a\overline w$. By factorizing the terms, we would obtain $r^2 - \lvert a \rvert^2=(1-r^2\lvert a\rvert^2)w\overline w+\overline aw(1-r^2)+a\overline w(1-r^2)$.
Manipulating the expression (as he wanted to sue completing the square method), he will obtain $(w+\frac{1-r^2}{1-r^2|a|^2}a)(\overline w+\frac{1-r^2}{1-r^2|a|^2}\overline a)=\frac{(r^2-|a|^2)(1-r^2|a|^2)+(1-r^2)^2|a|^2}{(1-r^2|a|^2)^2}$. $(*)$
Now just by focusing on the numerator in the right hand side of the equation;
$(r^2-|a|^2)(1-r^2|a|^2)+(1-r^2)^2|a|^2=r^2-r^4|a|^2-|a|^2+r^2|a|^4+(1-2r^2+r^4)|a|^2=r^2-r^4|a|^2-|a|^2+r^2|a|^4+|a|^2-2r^2|a|^2+|a|^2r^4=r^2-2r^2|a|^2+r^2|a|^4=r^2(1-2|a|^2+|a|^4=r^2(1-|a|^2)^2$
Plucking this expression back into the equation $(*)$, we have that
$\lvert w-(\frac{r^2-1}{1-r^2|a|^2}a) \rvert^2=\frac{r^2(1-|a|^2)^2}{(1-r^2|a|^2)^2}=(\frac{r(1-|a|^2}{1-r^2|a|^2})^2$. We can take away the squares on both sides to obtain
$\lvert w-\frac{r^2-1}{1-r^2|a|^2}a \rvert=\frac {r(1-|a|^2)}{1-r^2|a|^2}$. Then he gave a conclusion that because $0 \leq \lvert \frac{1-r^2}{1-r^2|a|^2}\rvert |a|+\frac{r(1-|a|^2)}{1-r^2|a|^2} \leq 1$, then it must be that $f_a(z)$ maps the unit disk $D[0,1]$ to itself in a bijective fashion.
I would like to ask if there are any other shorter and plausible methods in doing part (c) as I am super confused as to why he is doing such a longer method. He says that this method is one method he can think of, so I was wondering if there are indeed other methods to do so. I hope you can help me understand Complex Analysis better! Thanks so much for your help!!
There is a unique Möbius transformation mapping a given set of 3 points in the extended complex plane $\mathbb{C} \cup \{\infty\}$ to another set of 3 points.
You can use that fact to show that your function maps the unit circle to itself. It might map the unit disk to the set of points outside the unit circle, but you can show that it doesn't by looking at the image of any point in the disk, say $0$.
By b), $f_a$ has an inverse, so it must be bijective. The result follows.