In Stein's Complex Analysis, he presents the following statement and proof of Casorati-Weierstrass:
Suppose $f$ is holomorphic in the punctured disc $D_r(z_0) – \{z_0\}$ and has an essential singularity at $z_0$. Then, the image of $d_r(z_0) – \{z_0\}$ under $f$ is dense in the complex plane.
Proof:
We argue by contradiction. Assume that the range of $f$ is not dense, so that there exists $w \in \mathbb{C}$ and $\delta > 0$ such that
$$|f(z) – w| > \delta$$
for all $z \in D_r(z_0) – \{z_0\}$. We may therefore define a new function on $D_r(z_0) – \{z_0\}$ by
$$g(z) = \frac{1}{f(z) – w}$$
which is holomorphic on the punctured disc and bounded by $1/\delta$. Hence, $g$ has a removable singularity at $z_0$.
Statement 1: If $g(z_0) \neq 0$, then $f(z) – w$ is holomorphic at $z_0$, which contradicts the assumption that $z_0$ is an essential singularity.
Statement 2: In the case that $g(z_0) = 0$, then $f(z) – w$ has a pole at $z_0$ also contradicting the nature of the singularity at $z_0$. The proof is complete.
My Question:
In regards to statement 1, why precisely is this true? I can rewrite $g(z) = \frac{1}{f(z) – w}$ as $f(z) – w = \frac{1}{g(z)}$, and if $g(z_0) \neq 0$, then $f(z_0) – w$ is defined — does it immediately follow that $f(z_0)$ is holomorphic at $z_0$?
Best Answer
Your reasoning is pretty much correct. Since $f$ is holomorphic on $D_r(z_0)\setminus\{z_0\}$, and $\lim_{z\to z_0}f(z)=\frac{1}{g(z_0)}-w$ exists, it follows the singularity at $z_0$ is removable, contradicting that it is essential. This is one of the many ways to tell if a singularity is removable.
This is what they mean when they say $f$ is holomorphic at $z_0$; they mean the singularity is removable there, so it can be made holomorphic by an appropriate definition of $f(z_0)$.