Consider the complex mapping defined by $$f_a(z)=\frac{z-a}{\bar az-1}$$ for some constant $a.$ I want to show that $f$ maps the unit circle to itself. I know that the unit circle is defined by $|z|=1,$ or equivalently $z=e^{i \theta}.$ This gives me $$f_a(z)=f_a(e^{i \theta})=\frac{e^{i \theta}-a}{\bar a e^{i \theta}-1}$$
I'm not sure how to proceed from here though.
Additional question: I also want to show that of $a$ lies inside the unit disk then $f$ maps the unit disk to itself. In order to answer this I have shown that $$|\bar a z-1|^2 – |z-a|^2 =(1-|a|^2)(1-|z|^2)<1$$
How am I able to use this to deduce the result?
Best Answer
Hint. Note that $\overline{e^{i \theta}-a}= -e^{-i\theta}(\overline{a}e^{i\theta}-1)$ implies that $|e^{i \theta}-a|=|\overline{a}e^{i\theta}-1|$.
P.S. As regards your additional question, if $|z|<1$ and $|a|<1$ then $$|\bar a z-1|^2-|z-a|^2=(\bar a z-1)( a \bar z-1)-(z-a)(\bar z-\bar a)=|a|^2|z|^2+1-|z|^2-|a|^2\\=(1-|a|^2)(1-|z|^2).$$ Hence (note that $\bar a z\not=1$), $$1 - \frac{|z-a|^2}{|\bar a z-1|^2} =\frac{(1-|a|^2)(1-|z|^2)}{|\bar a z-1|^2}>0$$ which implies that $$\left|\frac{z-a}{\bar a z-1}\right|^2<1$$ that is $|f(z)|<1$.