[proofplan] We establish the [Laurent series expansion](/theorems/350) by applying the one-variable [Cauchy integral formula](/theorems/345) iteratively on a torus $\mathbb{T}^n(R)$ for radii $R_j > |z_j|$ chosen so that $\overline{\mathbb{D}^n(0, R)} \subset \Omega$. The iterated Cauchy formula represents $f(z)$ as an $n$-fold contour integral over the distinguished boundary. Expanding each Cauchy kernel $(\zeta_j - z_j)^{-1}$ as a geometric series in $z_j/\zeta_j$ produces the [power series](/page/Power%20Series) with coefficients given by integral formulas over the torus. Since the domain is a complete Reinhardt domain, the origin belongs to $\Omega$ and no negative powers appear. Absolute and [uniform convergence](/page/Uniform%20Convergence) on compact subsets follows from the [Cauchy estimates](/theorems/2571). Uniqueness of the coefficients follows from the orthogonality of monomials under torus integration. [/proofplan] [step:Represent $f(z)$ via the iterated Cauchy integral formula on a torus $\mathbb{T}^n(R)$] Fix $z \in \Omega \cap (\mathbb{C}^*)^n$ and set $r_j = |z_j| > 0$ for $j = 1, \dots, n$. Since $\Omega$ is a complete Reinhardt domain and open, there exist $R_j > r_j$ such that $\overline{\mathbb{D}^n(0, R)} \subset \Omega$. To verify this: since $z \in \Omega$ and $\Omega$ is open, there exists $\varepsilon > 0$ with $B(z, \varepsilon) \subset \Omega$. Choose $R_j = r_j + \varepsilon / \sqrt{n}$; then any $w$ with $|w_j| \leq R_j$ for all $j$ satisfies $|w - z| \leq \sqrt{n} \cdot \varepsilon/\sqrt{n} = \varepsilon$ when each $w_j$ has the same phase as $z_j$, and by completeness of the Reinhardt domain (if $w' \in \Omega$ then every $w''$ with $|w''_j| \leq |w'_j|$ belongs to $\Omega$), we get $\overline{\mathbb{D}^n(0, R)} \subset \Omega$. The [Cauchy Integral Formula on Polydiscs](/theorems/3398) applied to the polydisc $\mathbb{D}^n(0, R)$ gives \begin{align*} f(z) = \frac{1}{(2\pi i)^n} \oint_{|\zeta_1| = R_1} \cdots \oint_{|\zeta_n| = R_n} \frac{f(\zeta)}{(\zeta_1 - z_1) \cdots (\zeta_n - z_n)}\, d\zeta_n \cdots d\zeta_1, \end{align*} which is valid since $|z_j| = r_j < R_j$ for each $j$. [guided] Why do we need to enlarge the radii from $r_j$ to $R_j$? The polydisc Cauchy formula requires the point $z$ to lie strictly inside the polydisc, i.e., $|z_j| < R_j$ for all $j$. Since $|z_j| = r_j$, we cannot use the torus at radii $r_j$ directly. Instead, we choose $R_j > r_j$ close enough to $r_j$ that $\overline{\mathbb{D}^n(0, R)} \subset \Omega$. Such $R$ exists because $\Omega$ is open and contains $z$. The completeness of the Reinhardt domain ensures that $\overline{\mathbb{D}^n(0, R)} \subset \Omega$: if a point $w$ with $|w_j| = R_j$ belongs to $\Omega$, then every $w'$ with $|w'_j| \leq R_j$ also belongs to $\Omega$. Fixing $z \in \Omega \cap (\mathbb{C}^*)^n$ with $r_j = |z_j| > 0$, we choose $R_j > r_j$ so that the closed polydisc $\overline{\mathbb{D}^n(0, R)}$ lies in $\Omega$. The [Cauchy Integral Formula on Polydiscs](/theorems/3398) then applies to $z \in \mathbb{D}^n(0, R)$, giving \begin{align*} f(z) = \frac{1}{(2\pi i)^n} \oint_{|\zeta_1| = R_1} \cdots \oint_{|\zeta_n| = R_n} \frac{f(\zeta)}{(\zeta_1 - z_1) \cdots (\zeta_n - z_n)}\, d\zeta_n \cdots d\zeta_1. \end{align*} The hypothesis of the polydisc Cauchy formula is that $z$ lies in the open polydisc $\mathbb{D}^n(0, R)$, which holds since $|z_j| = r_j < R_j$ for each $j = 1, \dots, n$. [/guided] [/step] [step:Expand each Cauchy kernel as a geometric series to obtain the power series] For each $j = 1, \dots, n$, since $|\zeta_j| = R_j > |z_j| = r_j$, the Cauchy kernel admits the geometric series expansion \begin{align*} \frac{1}{\zeta_j - z_j} = \frac{1}{\zeta_j} \cdot \frac{1}{1 - z_j/\zeta_j} = \sum_{\alpha_j = 0}^{\infty} \frac{z_j^{\alpha_j}}{\zeta_j^{\alpha_j + 1}}, \end{align*} converging absolutely and uniformly in $\zeta_j$ on $\{|\zeta_j| = R_j\}$ since $|z_j/\zeta_j| = r_j/R_j < 1$. Taking the product over all $j = 1, \dots, n$: \begin{align*} \prod_{j=1}^n \frac{1}{\zeta_j - z_j} = \sum_{\alpha \in \mathbb{N}_0^n} \frac{z^\alpha}{\zeta^{\alpha + \mathbf{1}}} \end{align*} where $\alpha = (\alpha_1, \dots, \alpha_n) \in \mathbb{N}_0^n := \{0, 1, 2, \dots\}^n$, $z^\alpha = z_1^{\alpha_1} \cdots z_n^{\alpha_n}$, $\zeta^{\alpha + \mathbf{1}} = \zeta_1^{\alpha_1 + 1} \cdots \zeta_n^{\alpha_n + 1}$, and $\mathbf{1} = (1, \dots, 1)$. The product series converges absolutely and uniformly on the torus $\{|\zeta_j| = R_j\}$. Substituting into the Cauchy integral and interchanging summation with integration (justified by the [uniform convergence](/page/Uniform%20Convergence) of the geometric series on the compact torus): \begin{align*} f(z) &= \frac{1}{(2\pi i)^n} \sum_{\alpha \in \mathbb{N}_0^n} z^\alpha \oint_{|\zeta_1| = R_1} \cdots \oint_{|\zeta_n| = R_n} \frac{f(\zeta)}{\zeta^{\alpha + \mathbf{1}}}\, d\zeta_n \cdots d\zeta_1. \end{align*} This gives the [power series](/page/Power%20Series) expansion $f(z) = \sum_{\alpha \in \mathbb{N}_0^n} a_\alpha z^\alpha$ with coefficients \begin{align*} a_\alpha = \frac{1}{(2\pi i)^n} \oint_{|\zeta_1| = R_1} \cdots \oint_{|\zeta_n| = R_n} \frac{f(\zeta)}{\zeta^{\alpha + \mathbf{1}}}\, d\zeta_n \cdots d\zeta_1. \end{align*} [guided] The expansion uses the standard technique of expressing the Cauchy kernel as a geometric series and interchanging sum and integral. For each $j = 1, \dots, n$, on the contour $|\zeta_j| = R_j$ with $|z_j| = r_j < R_j$, the ratio $|z_j/\zeta_j| = r_j/R_j < 1$, so the geometric series \begin{align*} \frac{1}{\zeta_j - z_j} = \frac{1}{\zeta_j} \cdot \frac{1}{1 - z_j/\zeta_j} = \sum_{\alpha_j = 0}^{\infty} \frac{z_j^{\alpha_j}}{\zeta_j^{\alpha_j + 1}} \end{align*} converges absolutely and uniformly in $\zeta_j$ on $\{|\zeta_j| = R_j\}$. The product of $n$ such series gives \begin{align*} \prod_{j=1}^n \frac{1}{\zeta_j - z_j} = \sum_{\alpha \in \mathbb{N}_0^n} \frac{z^\alpha}{\zeta^{\alpha + \mathbf{1}}}, \end{align*} where the multi-index notation $z^\alpha = z_1^{\alpha_1} \cdots z_n^{\alpha_n}$, $\zeta^{\alpha + \mathbf{1}} = \zeta_1^{\alpha_1 + 1} \cdots \zeta_n^{\alpha_n + 1}$, and $\mathbf{1} = (1, \dots, 1)$. This product series converges absolutely and uniformly on the torus $\mathbb{T}^n(R) = \{|\zeta_j| = R_j\}$ because each factor converges uniformly and the product of finitely many uniformly convergent series converges uniformly. Why can we interchange sum and integral? The integrand $f(\zeta) / (\zeta_1 - z_1) \cdots (\zeta_n - z_n)$ is replaced by the series $f(\zeta) \sum_\alpha z^\alpha / \zeta^{\alpha + \mathbf{1}}$. Since $f$ is continuous on the compact torus $\mathbb{T}^n(R)$, it is bounded there by some $M_R = \sup_{\mathbb{T}^n(R)} |f| < \infty$. The partial sums of the series are bounded by $M_R \sum_\alpha (r/R)^\alpha$, which converges (product of $n$ geometric series with ratio $r_j/R_j < 1$). By the Weierstrass $M$-test, the interchange is justified, giving \begin{align*} f(z) = \sum_{\alpha \in \mathbb{N}_0^n} a_\alpha z^\alpha, \qquad a_\alpha = \frac{1}{(2\pi i)^n} \oint_{\mathbb{T}^n(R)} \frac{f(\zeta)}{\zeta^{\alpha + \mathbf{1}}}\, d\zeta. \end{align*} The expansion above produces a [power series](/page/Power%20Series) in non-negative multi-indices $\alpha \in \mathbb{N}_0^n$. But the theorem claims a [Laurent series](/page/Laurent%20Series) with $\alpha \in \mathbb{Z}^n$. Where do the negative powers come from? On a complete Reinhardt domain containing the origin (which is guaranteed by completeness: if $z \in \Omega$ then the entire closed polydisc $\{|w_j| \leq |z_j|\}$ is in $\Omega$, including the origin), the function $f$ is holomorphic at $0$ and the expansion is a Taylor series -- no negative powers appear. The [Laurent series](/page/Laurent%20Series) statement with $\alpha \in \mathbb{Z}^n$ reduces to a Taylor series with $a_\alpha = 0$ for $\alpha \notin \mathbb{N}_0^n$. [/guided] [/step] [step:Extend to a Laurent series with $\alpha \in \mathbb{Z}^n$ and establish independence of radii] For a complete Reinhardt domain $\Omega$ and $f \in \mathcal{O}(\Omega)$, define the Laurent coefficients for all $\alpha \in \mathbb{Z}^n$ by \begin{align*} a_\alpha = \frac{1}{(2\pi i)^n} \oint_{|\zeta_1| = \rho_1} \cdots \oint_{|\zeta_n| = \rho_n} \frac{f(\zeta)}{\zeta^{\alpha + \mathbf{1}}}\, d\zeta_n \cdots d\zeta_1 \end{align*} for any $\rho = (\rho_1, \dots, \rho_n)$ with $\mathbb{T}^n(\rho) \subset \Omega$ and $\rho_j > 0$. The coefficient $a_\alpha$ is independent of the choice of $\rho$: for any two valid radii $\rho, \rho'$, the integrand $f(\zeta)/\zeta^{\alpha + \mathbf{1}}$ is holomorphic in each $\zeta_j$ on the annulus $\{\min(\rho_j, \rho_j') < |\zeta_j| < \max(\rho_j, \rho_j')\}$ (with the other variables fixed on the torus), so the one-variable Cauchy theorem gives independence of each radius. Since $\Omega$ is a complete Reinhardt domain, the origin belongs to $\Omega$ (take any $w \in \Omega$ and apply completeness: $0 \in \Omega$ since $|0| \leq |w_j|$ for all $j$). Therefore $f$ is holomorphic at the origin, and the Taylor expansion from the previous step gives $f(z) = \sum_{\alpha \in \mathbb{N}_0^n} a_\alpha z^\alpha$. The coefficients $a_\alpha = 0$ for all $\alpha \in \mathbb{Z}^n \setminus \mathbb{N}_0^n$, so the [Laurent series](/page/Laurent%20Series) reduces to a Taylor series: \begin{align*} f(z) = \sum_{\alpha \in \mathbb{Z}^n} a_\alpha z^\alpha = \sum_{\alpha \in \mathbb{N}_0^n} a_\alpha z^\alpha. \end{align*} [guided] For a complete Reinhardt domain $\Omega$ and $f \in \mathcal{O}(\Omega)$, we define the Laurent coefficients for all $\alpha \in \mathbb{Z}^n$ by \begin{align*} a_\alpha = \frac{1}{(2\pi i)^n} \oint_{|\zeta_1| = \rho_1} \cdots \oint_{|\zeta_n| = \rho_n} \frac{f(\zeta)}{\zeta^{\alpha + \mathbf{1}}}\, d\zeta_n \cdots d\zeta_1 \end{align*} for any $\rho = (\rho_1, \dots, \rho_n)$ with $\mathbb{T}^n(\rho) \subset \Omega$ and $\rho_j > 0$. Why is this well-defined, i.e., independent of the choice of $\rho$? For any two valid radii $\rho, \rho'$, the integrand $f(\zeta)/\zeta^{\alpha + \mathbf{1}}$ is holomorphic in each $\zeta_j$ on the annulus $\{\min(\rho_j, \rho_j') < |\zeta_j| < \max(\rho_j, \rho_j')\}$ (with the other variables fixed on the torus). The one-variable Cauchy theorem then gives independence of each radius, since a [holomorphic function](/page/Holomorphic%20Function) has the same contour integral over any two concentric circles bounding an annulus on which it is holomorphic. Now we show that the [Laurent series](/page/Laurent%20Series) reduces to a Taylor series. The crucial observation is that completeness of the Reinhardt domain forces the origin to lie in $\Omega$. Why? A complete Reinhardt domain satisfies: if $w \in \Omega$ and $|w'_j| \leq |w_j|$ for all $j = 1, \dots, n$, then $w' \in \Omega$. Taking $w' = 0$ (which satisfies $|0| \leq |w_j|$ for any $w \in \Omega$), we get $0 \in \Omega$. This means every [holomorphic function](/page/Holomorphic%20Function) on $\Omega$ is holomorphic at the origin, so no negative powers can appear in the expansion. More precisely, since $0 \in \Omega$ and $f$ is holomorphic there, the Taylor expansion from the previous step gives $f(z) = \sum_{\alpha \in \mathbb{N}_0^n} a_\alpha z^\alpha$ near the origin. By the [Identity Principle](/theorems/3357) (applied on the connected domain $\Omega$), this expansion holds on all of $\Omega \cap (\mathbb{C}^*)^n$. Therefore the coefficients $a_\alpha = 0$ for all $\alpha \in \mathbb{Z}^n \setminus \mathbb{N}_0^n$, and the [Laurent series](/page/Laurent%20Series) reduces to a Taylor series: \begin{align*} f(z) = \sum_{\alpha \in \mathbb{Z}^n} a_\alpha z^\alpha = \sum_{\alpha \in \mathbb{N}_0^n} a_\alpha z^\alpha. \end{align*} For a Reinhardt domain that is not complete (e.g., a polyannulus $\{r_j < |z_j| < R_j\}$), genuine negative powers can appear, but such domains are not complete Reinhardt domains. [/guided] [/step] [step:Establish absolute convergence on compact subsets of $\Omega \cap (\mathbb{C}^*)^n$] Let $L \Subset \Omega \cap (\mathbb{C}^*)^n$ be compact. Choose $R_j > \sup_{z \in L} |z_j|$ with $\overline{\mathbb{D}^n(0, R)} \subset \Omega$. Set $M_R = \sup_{\mathbb{T}^n(R)} |f| < \infty$ (continuous function on a compact set). The Cauchy estimate on the torus $\mathbb{T}^n(R)$ gives \begin{align*} |a_\alpha| = \left|\frac{1}{(2\pi i)^n} \oint_{\mathbb{T}^n(R)} \frac{f(\zeta)}{\zeta^{\alpha + \mathbf{1}}}\, d\zeta\right| \leq \frac{1}{(2\pi)^n} \cdot M_R \cdot \prod_{j=1}^n \frac{2\pi R_j}{R_j^{\alpha_j + 1}} = \frac{M_R}{R^\alpha}, \end{align*} where we used $|d\zeta_j| = R_j \, d\theta_j$ on the circle $|\zeta_j| = R_j$ and $|\zeta_j^{\alpha_j + 1}| = R_j^{\alpha_j + 1}$. For $z \in L$ with $|z_j| \leq r_j < R_j$ where $r_j := \sup_{z \in L} |z_j|$: \begin{align*} \sum_{\alpha \in \mathbb{N}_0^n} |a_\alpha| |z^\alpha| \leq M_R \sum_{\alpha \in \mathbb{N}_0^n} \prod_{j=1}^n \left(\frac{r_j}{R_j}\right)^{\alpha_j} = M_R \prod_{j=1}^n \frac{1}{1 - r_j/R_j} < \infty. \end{align*} Each factor $\sum_{\alpha_j = 0}^\infty (r_j/R_j)^{\alpha_j} = 1/(1 - r_j/R_j)$ is a convergent geometric series since $r_j/R_j < 1$, and the product of $n$ finite sums is finite. This shows absolute and [uniform convergence](/page/Uniform%20Convergence) on $L$. [guided] Let $L \Subset \Omega \cap (\mathbb{C}^*)^n$ be compact. We need to find an upper bound for $\sum_{\alpha \in \mathbb{N}_0^n} |a_\alpha| |z^\alpha|$ that is uniform on $L$. Choose $R_j > \sup_{z \in L} |z_j|$ with $\overline{\mathbb{D}^n(0, R)} \subset \Omega$ (possible by completeness of $\Omega$ and the compactness of $L$). Set $M_R = \sup_{\mathbb{T}^n(R)} |f| < \infty$. The coefficient formula gives the Cauchy estimate \begin{align*} |a_\alpha| = \left|\frac{1}{(2\pi i)^n} \oint_{\mathbb{T}^n(R)} \frac{f(\zeta)}{\zeta^{\alpha + \mathbf{1}}}\, d\zeta\right| \leq \frac{1}{(2\pi)^n} \cdot M_R \cdot \prod_{j=1}^n \frac{2\pi R_j}{R_j^{\alpha_j + 1}} = \frac{M_R}{R^\alpha}, \end{align*} where we used $|d\zeta_j| = R_j \, d\theta_j$ on the circle $|\zeta_j| = R_j$ and $|\zeta_j^{\alpha_j + 1}| = R_j^{\alpha_j + 1}$. Why does this bound hold? The integrand is bounded by $M_R / R^{\alpha + \mathbf{1}}$ on the torus, and the $n$-dimensional torus has total arc length $\prod_{j=1}^n 2\pi R_j$, giving the factor $M_R / R^\alpha$ after cancellation. Setting $r_j := \sup_{z \in L} |z_j| < R_j$, for any $z \in L$ we have $|z_j| \leq r_j$, so \begin{align*} \sum_{\alpha \in \mathbb{N}_0^n} |a_\alpha| |z^\alpha| \leq M_R \sum_{\alpha \in \mathbb{N}_0^n} \prod_{j=1}^n \left(\frac{r_j}{R_j}\right)^{\alpha_j} = M_R \prod_{j=1}^n \sum_{\alpha_j = 0}^\infty \left(\frac{r_j}{R_j}\right)^{\alpha_j} = M_R \prod_{j=1}^n \frac{1}{1 - r_j/R_j} < \infty. \end{align*} Each factor is a convergent geometric series since $r_j/R_j < 1$. The multi-index sum factorises into a product of one-dimensional geometric series because $\prod_{j=1}^n (r_j/R_j)^{\alpha_j}$ is a product of independent terms, each depending on a single index $\alpha_j$. This shows absolute and [uniform convergence](/page/Uniform%20Convergence) on $L$, completing the proof that $f(z) = \sum_{\alpha \in \mathbb{N}_0^n} a_\alpha z^\alpha$ converges absolutely and uniformly on compact subsets of $\Omega \cap (\mathbb{C}^*)^n$. [/guided] [/step] [step:Verify uniqueness of the Laurent coefficients] We show the coefficients $a_\alpha$ are uniquely determined by $f$. Suppose $f(z) = \sum_{\alpha \in \mathbb{Z}^n} b_\alpha z^\alpha$ is another [Laurent series](/page/Laurent%20Series) converging absolutely on $\Omega \cap (\mathbb{C}^*)^n$. Fix $\beta \in \mathbb{Z}^n$ and any $\rho$ with $\mathbb{T}^n(\rho) \subset \Omega$. Multiplying both sides by $z^{-\beta - \mathbf{1}}$ and integrating over $\mathbb{T}^n(\rho)$: \begin{align*} \frac{1}{(2\pi i)^n} \oint_{\mathbb{T}^n(\rho)} \frac{f(\zeta)}{\zeta^{\beta + \mathbf{1}}}\, d\zeta &= \sum_{\alpha \in \mathbb{Z}^n} b_\alpha \cdot \frac{1}{(2\pi i)^n} \oint_{\mathbb{T}^n(\rho)} \zeta^{\alpha - \beta - \mathbf{1}}\, d\zeta. \end{align*} The interchange of summation and integration is justified by the absolute convergence of the series on $\mathbb{T}^n(\rho)$. The torus integral evaluates to \begin{align*} \frac{1}{(2\pi i)^n} \oint_{\mathbb{T}^n(\rho)} \zeta^{\gamma}\, d\zeta = \begin{cases} 1 & \text{if } \gamma = -\mathbf{1}, \\ 0 & \text{otherwise}, \end{cases} \end{align*} for $\gamma \in \mathbb{Z}^n$, since each one-variable integral $\frac{1}{2\pi i}\oint_{|\zeta_j| = \rho_j} \zeta_j^{\gamma_j}\, d\zeta_j$ equals $1$ if $\gamma_j = -1$ and $0$ otherwise. Therefore $b_\beta = a_\beta$ for all $\beta \in \mathbb{Z}^n$, establishing uniqueness. [/step]