# The Casorati-Weierstrass Theorem

Theorem 1 (The Casorati-Weierstrass Theorem): Let $U \subseteq \mathbb{C}$ be open and let $z_0 \in U$. If $f$ is holomorphic on $U \setminus \{ z_0 \}$ and if $f$ has an essential singularity at $z_0$ then for any open set $V \subset U$ containing $z_0$, the image $f(V \setminus \{ z_0 \})$ is dense in $\mathbb{C}$. |

**Proof:**Suppose instead that $f(V \setminus \{ z_0 \})$ is NOT dense in $\mathbb{C}$. Then there exists a point $b \in \mathbb{C}$ and an $\epsilon > 0$ such that $f(V \setminus \{ z_0\}) \cap B(b, \epsilon) = \emptyset$. Equivalently, this means that if $z \in V \setminus \{ z_0 \}$ then:

(1)

\begin{align} |f(z) – b| \geq \epsilon \end{align}

- Define a new function $g$ on $V \setminus \{ z_0 \}$ by:

(2)

\begin{align} g(z) = \frac{1}{f(z) – b} \end{align}

- Note that $g$ is holomorphic on $V \setminus \{ z_0 \}$ since $f(z) \neq b$ on $V \setminus z_0$ and since $f$ is holomorphic on $V \setminus \{ z_0 \}$. Notice also that $g$ is bounded on $V \setminus \{ z_0 \}$ since for all $z \in V \setminus \{ z_0 \}$ we have that:

(3)

\begin{align} \quad |g(z)| = \frac{1}{|f(z) – b|} \leq \frac{1}{\epsilon} \end{align}

- So $g$ is holomorphic and bounded on $V \setminus \{ z_0 \}$, so $g$ has an isolated singularity at $z_0$. So there are two cases to consider. Either $g(z_0) \neq 0$ or $g$ has a zero of order $k$ at $z_0$. Before we look at these two cases, observe that we can write $f$ in the form:

(4)

\begin{align} f(z) = b + \frac{1}{g(z)} \end{align}

**Case 1:**Suppose that $g(z_0) \neq 0$. This is immediately a contradiction with the formula for $f$ above, because then $f$ would have a removable singularity, which is a contradiction since $f$ has an essential singularity at $z_0$.

**Case 2:**Suppose that $g$ has a zero of order $k$ at $z_0$. Then this implies with the formula for $f$ above that $f$ has a pole of order $k$ at $z_0$ – which is a contradiction since $f$ has an essential singularity at $z_0$. $\blacksquare$