# Non-constant, odd and entire is surjective

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?

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