Here's a question in complex analysis that I don't get, neither do I understand the first part of its answer given in the book. I would really appreciate some help.
Question:
Show that if $f(z)$ is an entire function, and there is a nonempty disk
such that $f(z)$ does not attain any values in the disk, then $f(z)$ is
constant.
Answer:
If $f$ does not attain values in the disk $|w – c| < \epsilon$, then
$1/(f – c)$ is bounded, hence constant by Liouville's theorem, and $f$
is constant.
Entire function: a function that is analytic on the entire complex plane.
Liouville's theorem: every bounded entire function is a constant.
Source of the question: Gamelin, Complex Analysis.
Happy New Year!
Best Answer
It means that for some disc $B(a,r)$ we have that
$$ f(\mathbb{C}) \cap B(a,r) = \emptyset$$
Then, for all $z \in \mathbb{C}$,
$$ | f(z) - a | \geq r$$
and you can consider its multiplicative inverse and apply Liouville.