I was reading a proof of the following theorem.
If $U$ is the open unit disc in $\mathbb{C}$, and $f:\mathbb{C}\to\mathbb{C}$ is analytic, then $d(f(x),f(z))<d(x,z)$ if $x\not=z$ (here $d$ is the hyperbolic metric). Also, if $\sup_{x\in U}|f(x)|<1$, then $f$ is a contraction and has a fixed point, since $(U,d)$ is a complete metric space.
This is an application of the Banach contraction mapping theorem, but I was wondering if one could also argue that since $\sup_{x\in U}|f(x)|=\delta<1$, we have that $B_{\delta}(0)\subset\mathbb{C}$ (the closed ball) and $f$ restricted to this set maps $f:B_{\delta}(0)\to B_{\delta}(0)$. Then since $f$ is continuous, it has a fixed point by Brower fixed point theorem.
Is this correct? I was wondering that it might not be because it would perhaps make the argument using the hyperbolic space less interesting. But maybe it's that the Banach contraction theorem is easy to prove, whereas the Brower fixed point theorem is hard to prove, so maybe you have to make up for this in the hyperbolic argument with some additional theorems about analytic maps. Any thoughts?