complex-analysis
I came across the following question:
For each $b \in \mathbb{D}$, construct an automorphism $\phi$ of the unit disk that is not the identity map such that $\phi(b)=b$.
I know that all automorphisms of the disk must have the form $e^{i\theta}\left(\frac{a-z}{1-\overline{a}z}\right)$ where $a \in \mathbb{D}$. The only map that I can see would fix $b$ is just a rotation by some multiple of $2\pi$. How should I proceed?
Fix $w^* \in \Bbb D$. The $(f_n)$ are automorphisms of the unit disk, therefore there exists a sequence $(z_n)$ in $\Bbb D$ with $f_n(z_n) = w^*$ for all $n$.
Case 1: $z_{n_k} \to z^* \in \Bbb D$ for some subsequence $(z_{n_k})$ of $(z_n)$. Then $f_{n_k}(z_{n_k}) \to f(z^*)$ which implies $f(z^*) = w^*$.
Case 2: $|z_n| \to 1$. Applying the Schwarz–Pick theorem to the inverse functions $f_n^{-1}$ gives $$ \left|\frac{z-z_n}{1-\overline{z_n} z} \right| \le \left|\frac{f_n(z)-w^*}{1-\overline{w^*} f_n(z)} \right| \, , $$ which can equivalently be written as $$ \frac{(1-|f_n(z)|^2)(1-|w^*|^2)}{|1-\overline{w^*} f_n(z)|^2} \le \frac{(1-|z|^2)(1-|z_n|^2)}{|1-\overline{z_n} z|^2} \, . $$ (Actually equality holds for automorphisms, but that is not needed here). For fixed $z$ and $n \to \infty$ the right-hand side converges to zero, and it follows that $|f(z)| = 1$ for all $z \in \Bbb D$. That is not possible for a non-constant holomorphic function.
So only the first case can occur, which proves that $f: \Bbb D \to \Bbb D$ is surjective.
Remark: The condition that the limit function $f$ is not constant is important, as the example $$ f_n(z) = \frac{z-(1- \frac 1n)}{1-(1-\frac 1n)z} $$ demonstrates. The $f_n$ are automorphisms of the unit disk with $f_n \to -1$ locally uniformly.
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:
Best Answer
Assume $b \ne 0$ as otherwise problem trivial.
Take $\psi_1(z)=\frac{z+b}{1+\bar b z}$. Clearly $\psi_1(0)=b$. Now take $\psi_2(z)=\alpha\frac{z-b}{1-\bar b z}, |\alpha|=1, \alpha \ne 1$. Clearly $\psi_2(b)=0$ so $\phi=\psi_1 \circ \psi_2$ satisfies $\phi(b)=b$
Since $\psi_2(0)=-\alpha b \ne -b$, $\phi(0) \ne 0$ so $\phi$ is not the identity