The Casorati-Weierstrass Theorem

Table of Contents

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}

(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 $$z_0$$ . $\blacksquare$

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