# Entire analytic automorphisms of C

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 $$V$$.