[proofplan] We prove the equivalence of three definitions of holomorphicity in several complex variables by establishing the cycle $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (2)$ and $(2) \Rightarrow (1)$. The implication $(1) \Rightarrow (2)$ uses the polydisc [Cauchy integral formula](/theorems/345) (valid for $C^1$ functions satisfying Cauchy--Riemann) to expand $f$ as a convergent [power series](/page/Power%20Series). The implication $(2) \Rightarrow (1)$ follows from term-by-term differentiation of the [power series](/page/Power%20Series). The implication $(2) \Rightarrow (3)$ is immediate since a convergent [power series](/page/Power%20Series) is separately holomorphic and locally bounded. The implication $(3) \Rightarrow (2)$ is [Osgood's Lemma](/theorems/3379). [/proofplan]

[step:$(1) \Rightarrow (2)$: Derive the power series expansion from the Cauchy--Riemann condition via the polydisc Cauchy formula]Assume $f \in C^1(\Omega)$ and $\partial_{\bar{z}_j} f = 0$ for all $j = 1, \dots, n$. Fix $a \in \Omega$ and choose $r = (r_1, \dots, r_n)$ with $r_j > 0$ such that $\overline{\mathbb{D}^n(a, r)} \subset \Omega$. Since $f$ is $C^1$ and satisfies the Cauchy--Riemann equations in each variable separately, for each fixed $(z_2, \dots, z_n)$ the function $z_1 \mapsto f(z_1, z_2, \dots, z_n)$ is holomorphic in $z_1$ (a $C^1$ function of one complex variable satisfying $\partial_{\bar{z}_1} f = 0$ is holomorphic by the classical equivalence). Similarly for each other variable. Therefore $f$ is separately holomorphic on $\Omega$. Since $f \in C^1(\Omega)$, it is in particular continuous, hence bounded on the compact set $\overline{\mathbb{D}^n(a, r)}$. By [Osgood's Lemma](/theorems/3379) (or directly by iterating the one-variable Cauchy formula as in the proof of the [Cauchy Integral Formula on Polydiscs](/theorems/3398)), $f$ satisfies the iterated Cauchy integral representation on $\mathbb{D}^n(a, r)$. Expanding the Cauchy kernel as a geometric series in each variable (as in the proof of [Osgood's Lemma](/theorems/3379)) yields a convergent [power series](/page/Power%20Series) $f(z) = \sum_{\alpha \in \mathbb{N}_0^n} c_\alpha (z - a)^\alpha$ on $\mathbb{D}^n(a, r)$.[/step]

[guided]We start from condition (1): assume $f \in C^1(\Omega)$ and $\partial_{\bar{z}_j} f = 0$ for all $j = 1, \dots, n$. The goal is to show that $f$ is locally representable by a convergent [power series](/page/Power%20Series). **Why does $C^1$ plus Cauchy--Riemann give separate holomorphicity?** Fix $a \in \Omega$ and choose $r = (r_1, \dots, r_n)$ with $r_j > 0$ such that $\overline{\mathbb{D}^n(a, r)} \subset \Omega$. For each fixed $(z_2, \dots, z_n)$, consider the function $z_1 \mapsto f(z_1, z_2, \dots, z_n)$. This is a $C^1$ function of one complex variable satisfying $\partial_{\bar{z}_1} f = 0$. In one complex variable, the classical Cauchy--Riemann theorem tells us that a $C^1$ function satisfying $\partial_{\bar{z}} f = 0$ is holomorphic --- the $C^1$ regularity is what allows us to write $f$ as the integral of its differential form and apply Green's theorem to establish the [Cauchy integral formula](/theorems/345). Since the same argument applies to each variable $z_j$ with the other variables held fixed, $f$ is separately holomorphic on $\Omega$. **Why do we need local boundedness, and how do we get it?** To pass from separate holomorphicity to a [power series](/page/Power%20Series) representation, we need to iterate the one-variable Cauchy formula across all $n$ variables. This iteration requires that $f$ be locally bounded (so that the iterated integrals converge and [Fubini's theorem](/theorems/2961) applies). Since $f \in C^1(\Omega)$, it is in particular continuous, hence bounded on the compact set $\overline{\mathbb{D}^n(a, r)}$. **Obtaining the [power series](/page/Power%20Series).** With separate holomorphicity and local boundedness in hand, we apply [Osgood's Lemma](/theorems/3379) (or equivalently, iterate the one-variable [Cauchy integral formula](/theorems/345) directly as in the proof of the [Cauchy Integral Formula on Polydiscs](/theorems/3398)). This gives the iterated Cauchy integral representation of $f$ on the polydisc $\mathbb{D}^n(a, r)$. We then expand the Cauchy kernel $\frac{1}{\zeta_j - z_j}$ as a geometric series $\sum_{\alpha_j = 0}^\infty \frac{(z_j - a_j)^{\alpha_j}}{(\zeta_j - a_j)^{\alpha_j + 1}}$ in each variable. Since the geometric series converges uniformly on $\overline{\mathbb{D}^n(a, r)}$ (because $|z_j - a_j| < r_j \leq |\zeta_j - a_j|$ for $\zeta_j$ on the boundary), we may interchange summation and integration to obtain the convergent [power series](/page/Power%20Series) \begin{align*} f(z) = \sum_{\alpha \in \mathbb{N}_0^n} c_\alpha (z - a)^\alpha \end{align*} on $\mathbb{D}^n(a, r)$, where the coefficients $c_\alpha$ are given by the iterated Cauchy integrals. This establishes condition (2). Alternatively, one can bypass [Osgood's Lemma](/theorems/3379) and derive the [Cauchy integral formula](/theorems/345) directly: the $C^1$ condition and Cauchy--Riemann equations allow an application of [Stokes' theorem](/theorems/1530) to pass from an integral over the polydisc boundary to the iterated contour integral. The result is the same.[/guided]

[step:$(2) \Rightarrow (1)$: A convergent power series is $C^1$ and satisfies Cauchy--Riemann] Assume $f$ is locally representable by a convergent [power series](/page/Power%20Series): for each $a \in \Omega$, there is a polydisc $\mathbb{D}^n(a, r)$ and coefficients $\{c_\alpha\}_{\alpha \in \mathbb{N}_0^n}$ such that $f(z) = \sum_\alpha c_\alpha (z - a)^\alpha$ with absolute convergence on $\mathbb{D}^n(a, r)$. A convergent [power series](/page/Power%20Series) in $n$ complex variables can be differentiated term by term: for each $j \in \{1, \dots, n\}$, \begin{align*} \frac{\partial f}{\partial z_j}(z) &= \sum_{\alpha : \alpha_j \geq 1} c_\alpha \cdot \alpha_j \cdot (z_j - a_j)^{\alpha_j - 1} \prod_{k \neq j} (z_k - a_k)^{\alpha_k}, \end{align*} which is again a convergent [power series](/page/Power%20Series) on $\mathbb{D}^n(a, r)$. In particular, $\partial_{z_j} f$ is continuous, so $f \in C^1(\Omega)$. For the Cauchy--Riemann condition, observe that each monomial $(z - a)^\alpha = (z_1 - a_1)^{\alpha_1} \cdots (z_n - a_n)^{\alpha_n}$ is a polynomial in $z_1, \dots, z_n$ alone (no $\bar{z}_j$ appears). Therefore $\partial_{\bar{z}_j} [(z - a)^\alpha] = 0$ for every $\alpha$ and every $j$. Since term-by-term differentiation with respect to $\bar{z}_j$ is justified by [uniform convergence](/page/Uniform%20Convergence) on compact subsets: \begin{align*} \partial_{\bar{z}_j} f(z) &= \sum_\alpha c_\alpha \cdot \partial_{\bar{z}_j} [(z - a)^\alpha] = 0. \end{align*} Hence $f$ satisfies condition (1). [/step]

[step:$(2) \Rightarrow (3)$: A convergent power series is separately holomorphic and locally bounded] Assume $f(z) = \sum_\alpha c_\alpha (z - a)^\alpha$ converges absolutely on $\mathbb{D}^n(a, r)$. Fix $j \in \{1, \dots, n\}$ and fix all variables except $z_j$. The resulting function of $z_j$ alone is a convergent [power series](/page/Power%20Series) in the single variable $z_j - a_j$ (with coefficients depending on the fixed values of the other variables), hence holomorphic in $z_j$. Since $j$ was arbitrary, $f$ is separately holomorphic. For local boundedness: on any compact sub-polydisc $\overline{\mathbb{D}^n(a, s)}$ with $0 < s_j < r_j$, the series converges absolutely and the partial sums are uniformly bounded: \begin{align*} |f(z)| \leq \sum_\alpha |c_\alpha| \prod_{j=1}^n s_j^{\alpha_j} < \infty, \end{align*} so $f$ is bounded on $\overline{\mathbb{D}^n(a, s)}$. Every point of $\Omega$ has such a polydisc neighbourhood, so $f$ is locally bounded on $\Omega$. [/step]

[step:$(3) \Rightarrow (2)$: Apply Osgood's Lemma to obtain the power series representation] Assume $f$ is separately holomorphic and locally bounded on $\Omega$. By [Osgood's Lemma](/theorems/3379), $f \in \mathcal{O}(\Omega)$. The proof of [Osgood's Lemma](/theorems/3379) constructs, for each $a \in \Omega$, a convergent [power series](/page/Power%20Series) $\sum_\alpha c_\alpha (z - a)^\alpha$ representing $f$ on a polydisc $\mathbb{D}^n(a, r) \subset \Omega$, via the iterated Cauchy integral and geometric series expansion of the Cauchy kernel. Therefore $f$ satisfies condition (2). This completes the cycle of implications, establishing the equivalence of all three conditions. [/step]