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Theorem. (Casorati-Weierstrass; Lang Complex analysis) Let $0$ be an essential singularity of the function $f$ and let $D$ be a disc centered at $0$ on which $f$ is holomorphic except at $0$. Let $U$ be the complement of $0$ in $D$. Then $f(U)$ is dense in the complex numbers.

Proof. Suppose there is $\alpha\in\Bbb C$ and a positive number $s>0$ such that $$|f(z)-\alpha|>s,\ \forall z\in U.$$ The function $$g(z) = \frac{1}{f(z)-\alpha}$$ is then holomorphic on $U$, and bounded on the disc $D$. Hence $0$ is a removable singularity of $g$, and $g$ may be extended to a holomorphic function on all of $D$. It then follows that $1/g(z)$ has at most a pole at $0$, which means that $f(z)-\alpha$ has at most a pole, contradicting the hypothesis that $f(z)$ has an essential singularity.

Question. Why can we conclude that $1/g(z)$ has at most a pole at $0$?

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When you remove the singularity at $0$ you will get a holomorphic function on the disk which is not identically $0$. If the value at $0$ is $0$ then $g$ has a zero of some finite order $N$. [Theorem 10.18 in Rudin's RCA]. This implies that $f(z)-\alpha=\frac 1 {g(z)}$ has a pole of order $N$ at $0$.

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  • $\begingroup$ Why $g$ has a zero of some finite order? $\endgroup$ Jan 20 at 6:34
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    $\begingroup$ Any analytic function vanishing at a point $c$ which is not identically $0$ has a zero of finite order at that point. For a proof write down the power series $\sum a_n(z-c)^{n}$ at the point and note that there is a smallest $N$ such that $a_n \neq 0$. @love_sodam $\endgroup$ Jan 20 at 6:38

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