We have already seen that every rotation $z \mapsto e^{i \theta} \cdot z$, for a fixed $\theta \in \mathbb{R}$, is an automorphism of the disk. For $a \in \mathbb{D}$ define
$$\varphi_{a}(z)=\frac{a-z}{1-\bar{a} z} $$Lemma 2.4. For every $a \in \mathbb{D}, \varphi_{a}$ is an automorphism of $\mathbb{D}$.
$$
I am trying to visualize the above lemma using $\alpha = \frac{i}{2}$ giving me $$\frac{(1 + 2 i z)}{(2 i – z)}$$
however the conformal map it produces doesn't seem to map the disk to the disk. Am I computing this formula wrong? Is the visualization linked actually an automorphisim of the disk?
http://davidbau.com/conformal/#(1%20%2B%202%20i%20z)%2F(2%20i%20-%20z)
Best Answer
In the hope a picture is worth a few hundred words, here's the image of a polar grid under your $\varphi$, with the unit circle in blue: