2
$\begingroup$

If $f$ is a non-constant, entire analytic function, which is odd, then it is also surjective, due to Picard's Little Theorem. Clearly, $f(0)=0$, and hence $0$ is not missed, and if $f(z)\ne a$, for all $z$, then $f(z)\ne-a$, for all $z$. So if $f$ is not onto, it misses at least two values, and Picard's Little Theorem implies that $f$ is constant.

Is it possible to show this without Picard's Little Theorem?

Otherwise, Is it possible to derive Picard's Little Theorem from this odd version of Picard's Little Theorem?

$\endgroup$
  • $\begingroup$ How do you prove it with Picard's little Theorem? $\endgroup$ – José Carlos Santos Aug 14 '17 at 12:59
  • $\begingroup$ @JoséCarlosSantos I have explained it in my modified question. $\endgroup$ – Yiorgos S. Smyrlis Aug 14 '17 at 13:03
  • 1
    $\begingroup$ Did you mean to say if $f(z) \ne a$ for any $z$ then $f(z) \color{red}{\ne} -a$ for any $z$? $\endgroup$ – tilper Aug 14 '17 at 13:05
  • $\begingroup$ @YiorgosS.Smyrlis I should have guessed it myself. Thanks. $\endgroup$ – José Carlos Santos Aug 14 '17 at 13:17
  • $\begingroup$ @tilper Of course. I've edited the question. $\endgroup$ – José Carlos Santos Aug 14 '17 at 13:17

Your Answer

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

Browse other questions tagged or ask your own question.