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I'm trying to solve this problem: I have an entire function $f$ which is not a polynomial. I have to prove that there exists a dense subset $\Omega \subset \mathbb{C}$ such that for every $\omega \in \Omega$ the equation $f(z)=w$ has infinitely many solutions.

If $f$ is not a polynomial it must have an essential singularity at infinity, so for every $R$, $f(|z|>R)$ is an open dense subset of the complex plane, but I don't know if this is useful (it's the only thing about 'density' I could think of!). Thanks for any help.

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You're almost there. Now just add in the Cassorati-Weierstrass Theorem: http://en.wikipedia.org/wiki/Casorati%E2%80%93Weierstrass_theorem

Cassorati-Weierstrass gives a solution $z_0$ in some neighborhood of $\infty$. Now, we can apply the theorem again in the neighborhood of $\infty$ given by $\{z:|z|>|z_0|\}$. Repeating this process, we get infinitely many solutions.

By the way, there's a strengthened version of this theorem, which states that $\Omega$ is either $\mathbb{C}$ or $\mathbb{C}$ with a single point removed. This is Picard's Theorem (not to be confused with Picard's Little Theorem). http://en.wikipedia.org/wiki/Picard%27s_great_theorem

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  • $\begingroup$ The part i cannot understand is how i can prove there are 'infintely many' solutions. If i fix an R, then i could take $\Omega=f(|z|>R)$. But then i only know that $\forall \omega \in \Omega \exists z$ s.t $|z|>R$ and f(z)=w... $\endgroup$ – balestrav Apr 15 '12 at 21:52
  • $\begingroup$ @balestrav: I've edited my answer to address this. In short, you should be looking at smaller and smaller neighborhoods in the domain of the function, not the range. $\endgroup$ – Brett Frankel Apr 15 '12 at 21:58
  • $\begingroup$ Thank you, i missed it for a little! $\endgroup$ – balestrav Apr 15 '12 at 22:04
  • $\begingroup$ I got a doubt: I took $\Omega$ and i find a solution $z_0$ in $|z|>R$. If I then consider $|z|>|z_0|$ I know its image is dense, but I don't know if this image is exactly $\Omega$, so continuing this process how can i be sure i don't erase the preceding solutions found (e.g maybe $\omega=f(z_0) \notin f(|z|>|z_0|)$) $\endgroup$ – balestrav Apr 16 '12 at 21:29
  • $\begingroup$ The problems is as follows. You need to fix some choice of $\omega$ in $\Omega$. Then for that choice of $\omega$, you find a $z_0$ with $f(z_0)=\omega$. Then you shrink your domain and find $z_1$ with $f(z_1)=\omega$, etc. So for each fixed $\omega$ there are infinitely many values $z$ such that $f(z)=\omega$. Note that the $z$'s may come from anywhere in the complex plane. $\endgroup$ – Brett Frankel Apr 17 '12 at 2:04

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