If a holomorphic function $f$ on the punctured unit disc satisfies that for some fixed $N$ $f(z)=w$ has at most $N$ solutions for any $w$ [duplicate]

Question

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• Proof that a certain entire function is a polynomial 1 answer

Show that $0$ is not an essential singularity.

This is a homework question. I know that there is a theorem saying that if $f$ has an essential singularity $a$, then for all but two $w$ in $\mathbb{C}$, in any neighborhood of $a$, $f$ attains the value $w$.

Using this theorem the question is quite simple, but I am not sure if I am meant to quote this theorem, because it was only mentioned in my lectures without a proof.

I wonder if this can be shown without using the theorem I quoted.

Answer 1

Assume that $f$ did have an essential singularity at $0$. Let $A_n=B_{\frac{1}{n}}(0)-\{0\}$ be the punctured disc of radius $\frac{1}{n}$ centered at $0$. Then, Casorati-Weierstrass and open mapping theorem implies that $f(A_n)$ is open and dense, and so by the Baire Category Theorem $\displaystyle \bigcap_{n=1}^{\infty}f(A_n)$ is dense, and so in in particular, not empty. Note though that by assumption if $w\in \mathbb{C}$ there exists finitely many $z\in\mathbb{D}-\{0\}$ such that $f(z)=w$ and thus we see that for large enough $n$ we have that $w\notin f(A_n)$. Thus, $\displaystyle w\notin \bigcap_{n=1}^{\infty}f(A_n)$ and so $\displaystyle \bigcap_{n=1}^{\infty}f(A_n)=\varnothing$. This is a contradiction.

NOTE: The above is actually a proof that if $f$ has an essential singularity at $0$ then it must take infinitely many values infinitely many times on any open neighborhood of $0$.