I have the following Theorem in my lecture notes:
If $f:D \to \mathbb{C}$ is holomorphic on $D-\{z_0\}$ where $D$ is open and connected and $z_0$ is an isolated singularity of f, then: $f$ has a pole at $z_0$ iff $\lim_{z \to z_0} |f(z)| = \infty$.
One implication is easy but I have issues proving the converse. I believe it has to do with Casorati-Weierstrass Theorem. That is, if $z_0$ was an essential singularity then for all $\varepsilon >0$, $f(B(z_0,\varepsilon)-\{z_0\})$ would be dense in $\mathbb{C}$, but I don't see how to get a contradiction from there. On top of that, I don't even see how $e^{1/z}$ isn't a contradiction to the theorem in the first place. Am I missing something really obvious, or do I miss some assumptions? Thank you for your help.
Best Answer
If $|f(z)| \to \infty$ as $z \to z_0$ then $f(B(z_0,\epsilon)\setminus \{z_0\})$ does not intersect the open unit disk if $\epsilon$ is small enough. Since any dense set has to intersect this ball it follows that $z_0$ is not an essential singularity.
The functio $e^{1/z}$ does not have the property $|f(z)| \to \infty$ as $z \to 0$. For example this function has modulus $1$ when $z =\frac 1 {2n\pi i}$.