0
$\begingroup$

$f$ is a holomorphic function defined on the whole complex plane and whose image is the complement of the open ball of radius $R$ centered at the origin. Then what can be said about $f$?

According to Liouville's theorem every bounded entire function is a constant. Here $f$ is unbounded then for this entire function $f$ what can I conclude?

$\endgroup$
  • $\begingroup$ By Little Picard, $f$ is also constant in this case. (Even that is overkill!) $\endgroup$ – Alex Provost Oct 13 '17 at 13:33
  • 3
    $\begingroup$ You're saying that $|f(z)|\geqslant r > 0$ for some $r$, so consider $1/f$. $\endgroup$ – Pedro Tamaroff Oct 13 '17 at 13:34
  • $\begingroup$ Yeah I got it! Thanks. $\endgroup$ – user251057 Oct 13 '17 at 13:35
1
$\begingroup$

You're saying that $|f(z)|\geqslant r > 0$ for some $r$, so consider $1/f$. More generally if $f$ misses an open ball $B(z_0,r)$ then consider $1/(f-z_0)$, which is constant by Liouville. The concise way to state this is that entire functions have dense image in $\mathbb C$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy