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.


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.


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!

  • $\begingroup$ "$f$ attains the value $v$" means $v$ is in the image of $f$, that is, $v=f(z)$ for some $z$. $\endgroup$ Dec 31, 2020 at 19:35

1 Answer 1


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.


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