[proofplan] We prove the [Hartogs Extension Theorem](/theorems/3401): a [holomorphic function](/page/Holomorphic%20Function) on the Hartogs figure $H(\varepsilon) \subset \mathbb{D}^n$ extends holomorphically to the full polydisc $\mathbb{D}^n$ when $n \geq 2$. The strategy is to define an extension via the Cauchy integral in the last variable $z_n$ over a fixed circle $\{|\zeta| = \rho\}$ with $\rho \in (1 - \varepsilon, 1)$. This integral produces a function $F$ that is holomorphic on all of $\mathbb{D}^{n-1} \times \{|z_n| < \rho\}$. On the overlap with $H(\varepsilon)$, we verify $F = f$ using the one-variable Cauchy formula, and conclude $F$ extends $f$ to the entire polydisc by the [Identity Principle](/theorems/3357). [/proofplan] [step:Recall the geometry of the Hartogs figure $H(\varepsilon)$] Write $z = (z', z_n) \in \mathbb{C}^{n-1} \times \mathbb{C}$ with $z' = (z_1, \ldots, z_{n-1})$. The Hartogs figure is \begin{align*} H(\varepsilon) = \{z \in \mathbb{D}^n : |z'| < \varepsilon \text{ or } 1 - \varepsilon < |z_n| < 1\}, \end{align*} where $|z'| = \max_{1 \leq j \leq n-1} |z_j|$ (the polydisc norm on $\mathbb{C}^{n-1}$) and $0 < \varepsilon < 1$. This set consists of two overlapping pieces: a "thin polydisc" $\{|z'| < \varepsilon\} \times \mathbb{D}$ and an "annular shell" $\mathbb{D}^{n-1} \times \{1 - \varepsilon < |z_n| < 1\}$. [/step] [step:Define the candidate extension via the Cauchy integral in $z_n$] Fix $\rho \in (1 - \varepsilon, 1)$. For any $z' \in \mathbb{D}^{n-1}$ and any $\zeta$ with $|\zeta| = \rho$, the point $(z', \zeta) \in H(\varepsilon)$: indeed, either $|z'| < \varepsilon$ (so $(z', \zeta)$ lies in the thin polydisc) or $|z'| \geq \varepsilon$, in which case $|\zeta| = \rho > 1 - \varepsilon$ means $(z', \zeta)$ lies in the annular shell. In both cases, $f(z', \zeta)$ is defined. Define \begin{align*} F: \mathbb{D}^{n-1} \times \{|z_n| < \rho\} &\to \mathbb{C} \\ (z', z_n) &\mapsto \frac{1}{2\pi i} \oint_{|\zeta| = \rho} \frac{f(z', \zeta)}{\zeta - z_n}\, d\zeta. \end{align*} [guided] The key observation making this definition valid is purely geometric: the circle $\{|\zeta| = \rho\}$ with $\rho \in (1 - \varepsilon, 1)$ lies in the annular region $\{1 - \varepsilon < |z_n| < 1\}$, so for every $z' \in \mathbb{D}^{n-1}$, the point $(z', \zeta)$ belongs to $H(\varepsilon)$ and $f(z', \zeta)$ is well-defined. Why do we use this particular contour? In one variable, the [Cauchy integral formula](/theorems/345) reconstructs a [holomorphic function](/page/Holomorphic%20Function) inside a disc from its values on the boundary circle. Here we apply this idea to the last variable $z_n$ while treating $z'$ as a parameter. The circle $|\zeta| = \rho$ serves as the "boundary" of the disc $\{|z_n| < \rho\}$, and the integral reconstructs $f$ inside this disc — but now the integral makes sense for all $z' \in \mathbb{D}^{n-1}$, not just for $|z'| < \varepsilon$. Define \begin{align*} F: \mathbb{D}^{n-1} \times \{|z_n| < \rho\} &\to \mathbb{C} \\ (z', z_n) &\mapsto \frac{1}{2\pi i} \oint_{|\zeta| = \rho} \frac{f(z', \zeta)}{\zeta - z_n}\, d\zeta. \end{align*} This is the candidate extension of $f$ to the region $\mathbb{D}^{n-1} \times \{|z_n| < \rho\}$. [/guided] [/step] [step:Verify that $F$ is holomorphic on $\mathbb{D}^{n-1} \times \{|z_n| < \rho\}$] The function $F$ is holomorphic in $z_n$ for each fixed $z'$: the integrand $(z', \zeta) \mapsto f(z', \zeta)/(\zeta - z_n)$ is a continuous function of $\zeta$ on the compact set $\{|\zeta| = \rho\}$, and $\zeta \mapsto 1/(\zeta - z_n)$ is holomorphic in $z_n$ for $|z_n| < \rho$. Differentiation under the integral sign in $z_n$ is justified by the uniform bound $|(\zeta - z_n)^{-1}| \leq (\rho - |z_n|)^{-1}$ on the contour. Similarly, $F$ is holomorphic in each variable $z_j$ ($1 \leq j \leq n-1$) for fixed $(z_1, \ldots, z_{j-1}, z_{j+1}, \ldots, z_n)$: since $f$ is holomorphic in $z_j$ on $H(\varepsilon)$, the integrand $f(z', \zeta)/(\zeta - z_n)$ is holomorphic in $z_j$, and differentiation under the integral sign is again justified by the compactness of the contour and continuity of the integrand. By the [Equivalence of Holomorphicity Definitions](/theorems/3399) (or directly by [Osgood's Lemma](/theorems/3379)), separate holomorphicity in each variable implies joint holomorphicity. Hence $F \in \mathcal{O}(\mathbb{D}^{n-1} \times \{|z_n| < \rho\})$. [/step] [step:Show $F = f$ on the overlap region and conclude by the identity principle] On the thin polydisc $\{|z'| < \varepsilon\} \times \{|z_n| < \rho\}$, both $F$ and $f$ are defined. For each fixed $z'$ with $|z'| < \varepsilon$, the slice $z_n \mapsto f(z', z_n)$ is holomorphic on the full disc $\{|z_n| < 1\}$ (since $(z', z_n) \in H(\varepsilon)$ for all $|z_n| < 1$ when $|z'| < \varepsilon$). By the one-variable [Cauchy integral formula](/theorems/345) applied to the [holomorphic function](/page/Holomorphic%20Function) $z_n \mapsto f(z', z_n)$ on $\{|z_n| < 1\}$ with the contour $|\zeta| = \rho < 1$: \begin{align*} f(z', z_n) = \frac{1}{2\pi i} \oint_{|\zeta| = \rho} \frac{f(z', \zeta)}{\zeta - z_n}\, d\zeta = F(z', z_n) \end{align*} for all $|z_n| < \rho$. Thus $F = f$ on the [open set](/page/Open%20Set) $\{|z'| < \varepsilon\} \times \{|z_n| < \rho\}$. Now $F$ is holomorphic on $\mathbb{D}^{n-1} \times \{|z_n| < \rho\}$ and agrees with $f$ on the annular shell $\mathbb{D}^{n-1} \times \{1 - \varepsilon < |z_n| < \rho\} \subset H(\varepsilon)$ (where both are defined), since we have already shown $F = f$ on the thin polydisc portion. On the annular shell, $f$ is directly defined by hypothesis, and $F = f$ there follows by the [Identity Principle](/theorems/3357) applied in each slice $z_n \mapsto F(z', z_n) - f(z', z_n)$, which is holomorphic and vanishes on $\{|z_n| < \rho\} \cap \{1 - \varepsilon < |z_n|\}$ (a non-empty [open set](/page/Open%20Set)). Define the extension $\tilde{f}: \mathbb{D}^n \to \mathbb{C}$ by \begin{align*} \tilde{f}(z) = \begin{cases} f(z) & \text{if } z \in H(\varepsilon), \\ F(z) & \text{if } z \in \mathbb{D}^{n-1} \times \{|z_n| < \rho\}. \end{cases} \end{align*} Since $F = f$ on the overlap $H(\varepsilon) \cap (\mathbb{D}^{n-1} \times \{|z_n| < \rho\})$, the function $\tilde{f}$ is well-defined. The union $H(\varepsilon) \cup (\mathbb{D}^{n-1} \times \{|z_n| < \rho\}) = \mathbb{D}^n$ (since every point $(z', z_n) \in \mathbb{D}^n$ with $|z_n| < \rho$ is covered by the second piece, and every point with $|z_n| \geq \rho > 1 - \varepsilon$ is in the annular shell of $H(\varepsilon)$). Hence $\tilde{f} \in \mathcal{O}(\mathbb{D}^n)$ and $\tilde{f}|_{H(\varepsilon)} = f$. [guided] Let us verify the covering claim carefully. Take any $(z', z_n) \in \mathbb{D}^n$. **Case 1:** $|z_n| < \rho$. Then $(z', z_n) \in \mathbb{D}^{n-1} \times \{|z_n| < \rho\}$, which is the domain of $F$. **Case 2:** $|z_n| \geq \rho$. Since $\rho > 1 - \varepsilon$, we have $|z_n| > 1 - \varepsilon$, and since $|z_n| < 1$ (because $(z', z_n) \in \mathbb{D}^n$), the point $(z', z_n)$ lies in the annular shell $\mathbb{D}^{n-1} \times \{1 - \varepsilon < |z_n| < 1\} \subset H(\varepsilon)$. So every point of $\mathbb{D}^n$ is in the domain of at least one of $f$ or $F$, and on the overlap both functions agree. The extension $\tilde{f}$ is holomorphic on each piece and hence on all of $\mathbb{D}^n$. This result is a striking consequence of $n \geq 2$: in one variable, a [holomorphic function](/page/Holomorphic%20Function) on an annulus $\{1 - \varepsilon < |z| < 1\}$ cannot in general be extended to the full disc (consider $z \mapsto 1/z$). The Hartogs extension phenomenon has no analogue in dimension one and is one of the foundational differences in several complex variables. [/guided] [/step]