can I say this is a rotation

Question

If $f:B(0,1)\to B(0,1)$ is a bijective holomorphic function such that $f^{-1}$ is also holomorphic.Can I say that $f$ is a rotation?

Answer 1

As stated in the comments, this is false. We can, however, describe all of the automorphisms of the unit disk using the Schwarz lemma. Note that this gives you conditions under which $f$ is a rotation - that is $f(0) = 0$ and $|f(z)| = |z|$ for some nonzero $z \in B(0, 1)$. There is another sort of automorphism of $B(0, 1)$ defined as follows: let $a \in B(0, 1)$ and define $f_a(z) = \frac{a - z}{1 - \overline{a} z}$. This is an automorphism of $B(0, 1)$. This requires some checking using the maximum modulus principle to conclude that it actually maps into $B(0, 1)$ and the Schwarz lemma to conclude that it is its own inverse. We claim that every such automorphisms looks like $e^{i \theta} f_a$ for some $\theta, a$.

This proof is taken from Stein and Shakarchi's Complex Analysis Theorem 2.2. Let $f: B(0, 1) \longrightarrow B(0, 1)$ be an automorphism. Then there is some $a \in B(0, 1)$ such that $f(a) = 0$. Consider now $g = f \circ f_a$ and observe that $g(0) = f(f_a(0)) = f(a) = 0$. Then by the Schwarz lemma, $|g(z)| \leq |z|$. Furthermore, applying the Schwarz lemma to $g^{-1}$ yields $|g^{-1}(z)| \leq |z|$, so $|z| = |g^{-1}(g(z))| \leq |g(z)|$ so $|g(z)| = |z|$. Hence, $g$ is a rotation (again by the Schwarz lemma). Recall that we had $f \circ f_a = g$. As $f_a = f_a^{-1}$, $f = g \circ f_a$ as desired.