Casorati-Weierstraß theorem - dense

Question

I try to understand the Casorati-Weierstraß theorem. But I don't understand when a picture is dense in C. $e^{1/z}$ is dense, $1/z$ isn't. But why?

Thanks.

Answer 1

No function is (or isn't) dense. That theorem says that if $V$ is a neighborhood of an essential singularity at $a$ of a function $f$, then $f(V\setminus\{a\})$ is dense. Well, $0$ is not an essential singularity of $\frac1z$ (it's a simple pole), but it is an essential singularity of $\exp\left(\frac1z\right)$.