Prove that if $z \in\mathbb C$ which $|z|=1$ then $\displaystyle\left\vert{\frac{az+b}{\bar bz+\bar a}}\right\vert = 1 $ for $a,b \in \mathbb C$
Let $z \in\mathbb C$ such that $|z|=1$
Since, $|w|^2=w\bar w$ and $|z|=1 \implies |z|^2=1$
I've $$\left(\displaystyle \frac{az+b}{\bar bz+\bar a}\right)\left(\frac{\bar a\bar z+\bar b}{b\bar z +a}\right)=\frac{a\bar az\bar z+a\bar b z+\bar a b\bar z+b\bar b}{b\bar bz\bar z+a\bar b z+\bar ab \bar z + a\bar a}=\frac{|a|^2|z|^2+a\bar b z+\bar a b\bar z+|b|^2}{|b|^2|z|^2+a\bar b z+\bar a b\bar z+|a|^2}= \frac{|a|^2+a\bar b z+\bar a b\bar z+|b|^2}{|b|^2+a\bar b z+\bar a b\bar z+|a|^2}=1 $$
Hence $\displaystyle\left\vert{\frac{az+b}{\bar bz+\bar a}}\right\vert = 1 $ for $a,b \in \mathbb C$
Is my proof correct? Do you have another way?
Thank you.
Your proof is fine. A shorter proof is the following:
$$|z|=1 \Rightarrow z \bar{z}=1$$ Then $$\displaystyle\left\vert{\bar bz+\bar a}\right\vert=\displaystyle\left|\bar{z} \right|\left\vert{\bar bz+\bar a}\right\vert=\displaystyle\left\vert{\bar bz\bar{z}+\bar a \bar{z}}\right\vert=\displaystyle\left\vert{\bar b+\bar a \bar{z}}\right\vert=\displaystyle\left\vert{az+b}\right\vert$$