I was looking at a proof in Serge Lang's Introduction To Complex Analysis at a graduate-level regarding the form of analytic automorphisms which are entire functions I have a question about one of the steps in the proof.
I have trouble understanding how the assertion "$f$ is analytic automorphism of $\mathbf{C}$ implies that there exist $\delta, c> 0$ such that if $|w|>1\ \delta$ then $|f(w)>c|$" is made. If anybody could help me understand that, I would really appreciate it.
$f$ maps some neighbourhood $U$ of $0$ onto some neighbourhood $V$ of $0$. Now choose $\delta > 0$ and $ c > 0$ such that $U \subset B(0, 1/\delta) $ and $ B(0, c) \subset V$. Then $$ |w| > 1/\delta \implies w \notin U \underset{(*)}{\implies} f(w) \notin V \implies |f(w)| \ge c. $$
($B(z, r)$ denotes the open disk with center $z$ and radius $r$.)
The implication $(*)$ holds because $f$ is an automorphism, so that the complement of $U$ is mapped to the complement of $V$.
