[proofplan] We argue by contradiction. If the image of some punctured disk is not dense, then there exist $w \in \mathbb{C}$ and $\varepsilon > 0$ such that $|f(z) - w| \ge \varepsilon$ for all $z$ near $z_0$. The reciprocal $g(z) = 1/(f(z) - w)$ is then holomorphic and bounded on a punctured neighborhood of $z_0$, so by Riemann's Removable Singularity Theorem $g$ extends holomorphically across $z_0$. Analyzing the value $g(z_0)$ leads to $z_0$ being either a pole or a removable singularity of $f$, contradicting the hypothesis that $z_0$ is essential. [/proofplan] [step:Assume for contradiction that the image is not dense] Suppose, for the sake of contradiction, that there exist $r > 0$ with $B(z_0, r) \setminus \{z_0\} \subset \Omega$, a point $w \in \mathbb{C}$, and $\varepsilon > 0$ such that \begin{align*} |f(z) - w| \ge \varepsilon \quad \text{for all } z \in B(z_0, r) \setminus \{z_0\}. \end{align*} [/step] [step:Construct the bounded holomorphic function $g = 1/(f - w)$ on the punctured disk] Define \begin{align*} g: B(z_0, r) \setminus \{z_0\} &\to \mathbb{C} \\ z &\mapsto \frac{1}{f(z) - w}. \end{align*} Since $f \in \mathcal{O}(\Omega \setminus \{z_0\})$ and $f(z) - w \ne 0$ for $z \in B(z_0, r) \setminus \{z_0\}$ (as $|f(z) - w| \ge \varepsilon > 0$), $g$ is holomorphic on $B(z_0, r) \setminus \{z_0\}$. Moreover, $g$ is bounded: \begin{align*} |g(z)| = \frac{1}{|f(z) - w|} \le \frac{1}{\varepsilon} \quad \text{for all } z \in B(z_0, r) \setminus \{z_0\}. \end{align*} [guided] Why construct $g$? The assumption says the image of $f$ misses the open disk $B(w, \varepsilon)$. Taking the reciprocal of $f(z) - w$ converts "staying away from $w$" into a boundedness condition: if $|f(z) - w| \ge \varepsilon$, then $|1/(f(z) - w)| \le 1/\varepsilon$. Bounded holomorphic functions near isolated singularities are exactly the domain of Riemann's Removable Singularity Theorem. Formally, $g(z) = \frac{1}{f(z) - w}$ is well-defined and holomorphic on $B(z_0, r) \setminus \{z_0\}$ because $f(z) - w$ never vanishes there (its modulus is at least $\varepsilon$). The bound $|g(z)| \le 1/\varepsilon$ holds on the entire punctured disk. [/guided] [/step] [step:Apply Riemann's Removable Singularity Theorem to extend $g$ across $z_0$] Since $g$ is holomorphic and bounded on the punctured disk $B(z_0, r) \setminus \{z_0\}$, [Riemann's Removable Singularity Theorem](/theorems/3356) applies: $z_0$ is a removable singularity of $g$, and there exists a unique holomorphic extension $\tilde{g} \in \mathcal{O}(B(z_0, r))$ with $\tilde{g}(z) = g(z)$ for all $z \in B(z_0, r) \setminus \{z_0\}$. The hypotheses of Riemann's theorem are satisfied: $g \in \mathcal{O}(B(z_0, r) \setminus \{z_0\})$ and $g$ is bounded on $B(z_0, r) \setminus \{z_0\}$ (with bound $1/\varepsilon$). [/step] [step:Show that both cases $\tilde{g}(z_0) \ne 0$ and $\tilde{g}(z_0) = 0$ contradict essentiality] **Case 1: $\tilde{g}(z_0) \ne 0$.** Since $\tilde{g}$ is continuous and $\tilde{g}(z_0) \ne 0$, there exists $\delta > 0$ with $\tilde{g}(z) \ne 0$ for all $z \in B(z_0, \delta)$. On $B(z_0, \delta) \setminus \{z_0\}$: \begin{align*} f(z) = w + \frac{1}{g(z)} = w + \frac{1}{\tilde{g}(z)}. \end{align*} The function $z \mapsto w + 1/\tilde{g}(z)$ is holomorphic on $B(z_0, \delta)$ (since $\tilde{g}$ is holomorphic and non-vanishing there), and it agrees with $f$ on $B(z_0, \delta) \setminus \{z_0\}$. This means $z_0$ is a removable singularity of $f$, contradicting the hypothesis that $z_0$ is an essential singularity. **Case 2: $\tilde{g}(z_0) = 0$.** Since $\tilde{g}$ is holomorphic and not identically zero on $B(z_0, r)$ (because $|\tilde{g}(z)| = |g(z)| > 0$ for $z \ne z_0$ near $z_0$), the zero at $z_0$ has finite order: there exists $m \ge 1$ and a holomorphic function $h: B(z_0, r) \to \mathbb{C}$ with $h(z_0) \ne 0$ such that \begin{align*} \tilde{g}(z) = (z - z_0)^m h(z) \quad \text{for all } z \in B(z_0, r). \end{align*} Since $h(z_0) \ne 0$, there exists $\delta > 0$ with $h(z) \ne 0$ for $z \in B(z_0, \delta)$. On $B(z_0, \delta) \setminus \{z_0\}$: \begin{align*} f(z) = w + \frac{1}{\tilde{g}(z)} = w + \frac{1}{(z - z_0)^m h(z)}. \end{align*} The function $z \mapsto 1/h(z)$ is holomorphic and non-vanishing on $B(z_0, \delta)$, so $f(z) - w = (z - z_0)^{-m} / h(z)$ has a pole of order $m$ at $z_0$. This contradicts the hypothesis that $z_0$ is an essential singularity. [guided] We analyze the two exhaustive cases for $\tilde{g}(z_0)$. **If $\tilde{g}(z_0) \ne 0$:** then $1/\tilde{g}$ is holomorphic near $z_0$, so $f(z) = w + 1/\tilde{g}(z)$ extends holomorphically to $z_0$ with value $w + 1/\tilde{g}(z_0)$. This makes $z_0$ a removable singularity of $f$, which contradicts the hypothesis. **If $\tilde{g}(z_0) = 0$:** then $\tilde{g}$ has a zero of some finite order $m \ge 1$ at $z_0$. Why must $m$ be finite? Because $\tilde{g}$ is holomorphic and not identically zero (it equals $g \ne 0$ on the punctured disk), the zeros of a non-identically-zero holomorphic function are isolated and have finite order. We factor $\tilde{g}(z) = (z - z_0)^m h(z)$ with $h$ holomorphic and $h(z_0) \ne 0$. Then near $z_0$: \begin{align*} f(z) - w = \frac{1}{\tilde{g}(z)} = \frac{1}{(z - z_0)^m} \cdot \frac{1}{h(z)}. \end{align*} Since $1/h(z)$ is holomorphic and non-vanishing near $z_0$, this exhibits $f$ as having a pole of order $m$ at $z_0$. But a pole is not an essential singularity — contradiction. In both cases we reach a contradiction, so the original assumption that the image is not dense must be false. [/guided] [/step] [step:Conclude density of the image] Both cases lead to a contradiction with the hypothesis that $z_0$ is an essential singularity of $f$. Therefore the assumption that $f(B(z_0, r) \setminus \{z_0\})$ is not dense in $\mathbb{C}$ is false. That is, for every $w \in \mathbb{C}$ and every $\varepsilon > 0$, there exists $z \in B(z_0, r) \setminus \{z_0\}$ with $|f(z) - w| < \varepsilon$. [/step]