Show analytic isomorphism has no attracting fixed point

Question

Let $f: D\mapsto D$ analytic isomorphism for connected open subset $D \subset \mathbb{C}$. How do we show that $f$ has no attracting fixed point, i.e. $|f'(z)| \ge 1$ when $f(z) = z$?

Answer 1

If $D$ is hyperbolic, we can deduce it from the fact that $f$ is an isometry map for the hyperbolic metric on $D$. In other words, we have \begin{equation*} \forall z \in D,\ |f'(z)|\rho_{D}(f(z)) = \rho_{D}(z), \end{equation*} where $\rho_D$ denotes the hyperbolic density of $D$. It is well known that the hyperbolic density is positive. Thus, $f(z) = z$ implies $|f'(z)| = 1$.

If $D$ is not hyperbolic then its boundary contains at most two points. Julian Mejia gives you a counterexample for $D = \mathbb{C}$.

EDIT: my counterexample for $\mathbb{C}^{\star}$ was false.