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complex.pdf page52
| Theorem 8.10. (Casorati-Weierstrass): Let $U$ be an open subset of $\mathbb{C}$ and let $a \in U$. Suppose that $f: U \setminus\{a\} \to \mathbb{C}$ is a holomorphic function with an isolated essential singularity at $a$. Then for all $\rho>0$ with $B(a, \rho) \subseteq U$, the set $f(B(a, \rho) \setminus\{a\})$ is dense in $\mathbb{C}$, that is, the closure of $f(B(a, \rho) \setminus\{a\})$ is all of $\mathbb{C}$.
Proof.
Suppose, that there is some $\rho>0$ such that $z_0 \in \mathbb{C}$ is not a limit point of $f(B(a, \rho) \setminus\{a\})$.
Then the function $g(z)=1 /\left(f(z)-z_0\right)$ is bounded on $B(a, \rho) \setminus\{a\}$.
By Riemann's removable singularity theorem it extends to a holomorphic function on all of $B(a, \rho)$.
Since $f(z)=z_0+1 / g(z)$ if $g(a) \neq 0$ then $f(z)$ has a removable singularity at $a$.
If $g(a)=0,|1 / g(z)| \to \infty$ as $z \to a$, so $|f(z)| \to \infty$ as $z \to a$ and $f$ has a pole at $a$, a contradiction. $\square$
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Remark 8.11. In fact much more is true: Picard showed that if $f$ has an isolated essential singularity at $z_0$ then in any open disk about $z_0$ the function $f$ takes every complex value infinitely often with at most one exception. The example of the function $f(z)=\exp (1 / z)$, which has an essential singularity at $z=0$ shows that this result is best possible, since $f(z) \neq 0$ for all $z \neq 0$. |
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