Weierstrass-Casorati Theorem

Theorem
Let $f$ be an holomorphic function on $B \left({a, r}\right) \setminus \left\{{a}\right\}$.

If $f$ has an essential singularity at $a$, then $\forall s < r: f \left({B \left({a, s}\right) \setminus \left\{{a}\right\}}\right)$ is a dense subset of $\C$.

Proof
Without loss of generality, we can suppose $a=0$ and $r=1$.

Now, suppose that $\exists s < 1: f \left({B \left({0, s}\right) \setminus - \left\{{0}\right\}}\right)$ is not a dense subset of $\C$.

Then, by definition of dense subset:
 * $\exists z_0 \in \C: \exists r_0 > 0: B \left({z_0, r_0}\right) \cap f \left({B \left({0, s}\right) \setminus \left\{{0}\right\}}\right) = \varnothing$

Hence, the function $\varphi$ defined on $B \left({z_0, r_0}\right)$ by $\displaystyle \varphi \left({z}\right) = \frac 1 {f \left({z}\right) - z_0}$ is analytic on $B \left({0, s}\right) \setminus \left\{{0}\right\}$ and bounded near to $0$, because:
 * $\forall z \in B \left({0, s}\right) \setminus \left\{{0}\right\}: \left|{f \left({z}\right) - z_0}\right| > r_0 \implies \left|{\varphi \left({z}\right)}\right| < \frac 1 {r_0}$.

Therefore, we can extend the domain of $\varphi$ (using the Analytic Continuation Principle).


 * If $\varphi \left({0}\right) \ne 0$, then $\displaystyle f \left({0}\right) = z_0 + \frac 1 {\varphi \left({0}\right)}$ and the singularity of $f$ was removable.


 * Else, writing the power series of $\varphi$:
 * $\displaystyle \varphi \left({z}\right) = \sum_{n=1}^{+\infty} a_n z^n$

we see that:
 * $E = \left\{{k \in \N: a_k \neq 0}\right\} \neq \varnothing$

because $\varphi \neq 0$.

Putting $p = \min E$, we see that $0$ is a pole of order $p$ of $f$.

In each case, the assumption that $\exists s < 1: f \left({B \left({0, s}\right) \setminus \left\{{0}\right\}}\right)$ is not a dense subset of $\C$ contradicts the fact that $0$ is an essential singularity of $f$, which completes the proof.

It is also known as the Casorati-Weierstrass Theorem.