Answer: Casorati–Weierstrass: Dense Image Near Essential Singularity

Key Idea / Intuition

If the image of $f$ avoided a whole disk around some value $c$, then $1/(f(z)-c)$ would be bounded near $0$ — forcing $0$ to be either a removable singularity or a pole of $f$, contradicting the assumption that it's essential. So the image can never dodge an open set: it must be dense.

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Formal Proof / Solution

Theorem (Casorati–Weierstrass): If $f$ has an essential singularity at $z_0 = 0$, then for every punctured neighborhood $U = \{0 < |z| < r\}$, the image $f(U)$ is dense in $\mathbb{C}$.

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Proof by contradiction:

Suppose $f(U)$ is not dense in $\mathbb{C}$. Then there exists $c \in \mathbb{C}$ and $\varepsilon > 0$ such that

$|f(z) - c| \geq \varepsilon \quad \text{for all } z \in U.$

Define

$g(z) = \frac{1}{f(z) - c}, \quad z \in U.$

Since $|f(z) - c| \geq \varepsilon > 0$, we have $g$ holomorphic on $U$ and

$|g(z)| \leq \frac{1}{\varepsilon} \quad \text{for all } z \in U.$

So $g$ is bounded on the punctured disk. By Riemann's removable singularity theorem, $g$ extends to a holomorphic function $\tilde{g}$ on the full disk $|z| < r$.

Case 1: $\tilde{g}(0) \neq 0$.

Then $f(z) = c + 1/\tilde{g}(z)$ extends holomorphically to $z = 0$. So $z = 0$ is a removable singularity of $f$ — contradiction.

Case 2: $\tilde{g}(0) = 0$.

Then $\tilde{g}(z) = z^m h(z)$ near $0$, where $h(0) \neq 0$ and $m \geq 1$. Thus

$f(z) = c + \frac{1}{z^m h(z)},$

which has a pole of order $m$ at $z = 0$ — also a contradiction.

In both cases, we contradict the assumption that $z = 0$ is an essential singularity. $\blacksquare$

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Sanity check with $f(z) = e^{1/z}$:

Write $z = r e^{i\theta}$. Then $1/z = \frac{1}{r}e^{-i\theta}$, so

$e^{1/z} = e^{\cos\theta/r} \cdot e^{i\sin\theta/r}.$

• The modulus $e^{\cos\theta/r}$ ranges over $(0, \infty)$ as $\theta, r$ vary.
• The argument $\sin\theta/r$ ranges over all of $\mathbb{R}$ (mod $2\pi$).

So $e^{1/z}$ takes all nonzero values and its image is indeed dense in $\mathbb{C}$ (actually all of $\mathbb{C} \setminus \{0\}$, by Picard's great theorem — a much stronger result).

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Remark: Picard's Great Theorem strengthens this dramatically: near an essential singularity, $f$ takes every complex value with at most one exception, infinitely often. Casorati–Weierstrass is the elegant, accessible version of this idea.