TITLE: a question on transcendental entire function
QUESTION [5 upvotes]: $f$ is a transecendental entire funtion I need to show $\{w : f^{-1}(w) \text{ is infinite}\}$
is dense in $\mathbb{C}$
I have no idea how to prove it, please help.

REPLY [3 votes]: Here are the facts that you will need to know for the following (short proof):

Theorem: The only entire functions $f$ with a pole at infinity (i.e. such that $\displaystyle f\left(\frac{1}{z}\right)$ has a pole at $0$) are polynomials
Remark: You may know this result as "the only meromorphic functions on $\mathbb{P}^1$ are rational".

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Theorem(Open Mapping Theorem:) Every holomorphic function is an open mapping

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Theorem(Baire Category Theorem): In a complete metric space, the countable intersection of open dense subsets is dense.

You can find proofs for the first two in any standard complex analysis text, and the last in any standard point-set tex.

Ok, since $f$ is not a polynomial we know that $f$ has an essential singularity at $\infty$. In other words, $\displaystyle f\left(\frac{1}{z}\right)$ has an essential singularity at $0$ (this is the first theorem). Let $B_n'$ be the punctured disk of radius $\displaystyle \frac{1}{n}$ centered at $0$. Since $0$ is an essential singularity the Casorati-Weierstrass theorem tells us that each $f(B_n')$ is dense in $\mathbb{C}$. But, by the open mapping theorem we also know that $\displaystyle f(B_n')$ is open in $\mathbb{C}$. Thus, $\left\{f(B_n')\right\}$ is a countable collection of dense open subsets of $\mathbb{C}$, and thus by the Baire Category Theory $\displaystyle \bigcap_n f(B_n')$ is dense in $\mathbb{C}$. But, note that
$$\displaystyle \bigcap_n f(B_n')\subseteq\left\{w\in\mathbb{C}:f^{-1}(w)\text{ is infinite}\right\}$$
Indeed, if $\displaystyle w\in\bigcap_n f(B_n')$ then there exists a point $z_n$ in each $B_n'$ such that $f(z_n)=w$. Since any given point is in finitely many $B_n'$ we see that $\{z_n\}$ is infinite. Thus, $f^{-1}(w)$ is infinite.