[proofplan] The forward direction is immediate: for each $z_0 \notin \Omega$, the $1$-form $dz/(z - z_0)$ is a closed holomorphic $1$-form on $\Omega$, so its integral vanishes by the null-homologous hypothesis. The reverse direction uses de Rham cohomology: the winding number forms $dz/(z - z_j)$, where the $z_j$ are representatives of the bounded connected components of $\mathbb{C} \setminus \Omega$, form a basis for $H^1_{\mathrm{dR}}(\Omega)$. If all winding numbers vanish, then every closed $1$-form integrates to zero over $\gamma$, which is precisely the definition of null-homologous. [/proofplan] [step:Prove the forward direction: null-homologous implies vanishing winding numbers] Assume $\gamma$ is null-homologous in $\Omega$, meaning $\oint_\gamma \omega = 0$ for every closed holomorphic $1$-form $\omega$ on $\Omega$. Fix $z_0 \in \mathbb{C} \setminus \Omega$. Define the $1$-form \begin{align*} \omega_{z_0} := \frac{dz}{z - z_0}. \end{align*} Since $z_0 \notin \Omega$, the function $z \mapsto (z - z_0)^{-1}$ is holomorphic on $\Omega$, and $\omega_{z_0}$ is a closed holomorphic $1$-form on $\Omega$. By the null-homologous hypothesis: \begin{align*} n(\gamma, z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{dz}{z - z_0} = \frac{1}{2\pi i} \oint_\gamma \omega_{z_0} = 0. \end{align*} [guided] The forward direction is essentially a matter of recognising that the winding number is a special case of integrating a closed $1$-form. We want to show that if $\gamma$ is null-homologous (i.e., $\oint_\gamma \omega = 0$ for all closed holomorphic $1$-forms $\omega$ on $\Omega$), then $n(\gamma, z_0) = 0$ for every $z_0 \notin \Omega$. Fix $z_0 \in \mathbb{C} \setminus \Omega$. We define the $1$-form \begin{align*} \omega_{z_0} := \frac{dz}{z - z_0}. \end{align*} Why is this a closed holomorphic $1$-form on $\Omega$? The function $f: \Omega \to \mathbb{C}$ defined by $f(z) = (z - z_0)^{-1}$ is holomorphic on $\Omega$ because $z_0 \notin \Omega$, so $f$ has no singularities in $\Omega$. The $1$-form $\omega_{z_0} = f(z)\, dz$ is therefore holomorphic on $\Omega$. It is closed because every holomorphic $1$-form on a domain in $\mathbb{C}$ is closed (the Cauchy–Riemann equations imply $d\omega_{z_0} = 0$). Since $\omega_{z_0}$ is a closed holomorphic $1$-form on $\Omega$ and $\gamma$ is null-homologous, the hypothesis gives $\oint_\gamma \omega_{z_0} = 0$. Recalling the definition of the winding number: \begin{align*} n(\gamma, z_0) = \frac{1}{2\pi i} \oint_\gamma \frac{dz}{z - z_0} = \frac{1}{2\pi i} \oint_\gamma \omega_{z_0} = 0. \end{align*} The key point is that the winding number around $z_0$ is nothing but the period of a specific closed $1$-form, so the null-homologous condition (which kills all periods) automatically kills all winding numbers. [/guided] [/step] [step:Prove the reverse direction: vanishing winding numbers implies null-homologous] Assume $n(\gamma, z_0) = 0$ for all $z_0 \in \mathbb{C} \setminus \Omega$. We must show that $\oint_\gamma \omega = 0$ for every closed holomorphic $1$-form $\omega$ on $\Omega$. Let $\{C_j\}_{j \in J}$ denote the bounded connected components of $\mathbb{C} \setminus \Omega$, and choose a representative point $z_j \in C_j$ for each $j \in J$. Define \begin{align*} \omega_j := \frac{1}{2\pi i} \frac{dz}{z - z_j}, \quad j \in J. \end{align*} [claim:The cohomology classes $[\omega_j]$ form a basis for $H^1_{\mathrm{dR}}(\Omega)$] The first de Rham cohomology group $H^1_{\mathrm{dR}}(\Omega)$ of an open set $\Omega \subset \mathbb{C}$ is a free abelian group (over $\mathbb{C}$) with one generator for each bounded connected component of $\mathbb{C} \setminus \Omega$. The classes $[\omega_j]$, for $j \in J$, provide such a basis. [/claim] [proof] This follows from the topological structure of planar domains. The open set $\Omega$ deformation retracts onto a $1$-dimensional CW complex (a graph), so $H^1_{\mathrm{dR}}(\Omega) \cong H^1(\Omega; \mathbb{C})$ has dimension equal to the rank of $H_1(\Omega; \mathbb{Z})$, which equals the number of bounded connected components of $\mathbb{C} \setminus \Omega$. To see that the $[\omega_j]$ are linearly independent: for each bounded component $C_k$, choose a small circle $\sigma_k$ in $\Omega$ winding once around $z_k$. Then \begin{align*} \oint_{\sigma_k} \omega_j = \frac{1}{2\pi i} \oint_{\sigma_k} \frac{dz}{z - z_j} = n(\sigma_k, z_j) = \delta_{jk} \end{align*} where $\delta_{jk}$ is the Kronecker delta (since $\sigma_k$ winds once around $z_k$ and zero times around $z_j$ for $j \neq k$, as $z_j$ lies in a different connected component of $\mathbb{C} \setminus \Omega$ and hence in the unbounded component of $\mathbb{C} \setminus \operatorname{im}(\sigma_k)$ by the [Local Constancy of the Winding Number](/theorems/3360)). This duality pairing shows the $[\omega_j]$ are linearly independent. To see they span: any closed holomorphic $1$-form $\omega$ on $\Omega$ determines periods $a_j := \oint_{\sigma_j} \omega \in \mathbb{C}$. The form $\omega - \sum_{j \in J} a_j \cdot 2\pi i \cdot \omega_j$ has zero period around every generator $\sigma_j$ of $H_1(\Omega; \mathbb{Z})$, hence is exact on $\Omega$. Therefore $[\omega] = \sum_{j \in J} 2\pi i\, a_j\, [\omega_j]$, confirming that the $[\omega_j]$ span. [/proof] Now let $\omega$ be any closed holomorphic $1$-form on $\Omega$. By the basis decomposition, there exist constants $c_j \in \mathbb{C}$ and an exact $1$-form $df$ (where $f: \Omega \to \mathbb{C}$ is holomorphic) such that \begin{align*} \omega = \sum_{j \in J} c_j \cdot \frac{dz}{z - z_j} + df. \end{align*} Integrating over $\gamma$: \begin{align*} \oint_\gamma \omega = \sum_{j \in J} c_j \oint_\gamma \frac{dz}{z - z_j} + \oint_\gamma df = \sum_{j \in J} c_j \cdot 2\pi i \cdot n(\gamma, z_j) + 0. \end{align*} The exact form contributes zero since $\gamma$ is closed (so $\oint_\gamma df = f(\gamma(b)) - f(\gamma(a)) = 0$). Each term $n(\gamma, z_j) = 0$ by hypothesis (since $z_j \in C_j \subset \mathbb{C} \setminus \Omega$). Therefore $\oint_\gamma \omega = 0$. Since $\omega$ was an arbitrary closed holomorphic $1$-form on $\Omega$, the curve $\gamma$ is null-homologous in $\Omega$. [guided] The reverse direction is the substantial content of the theorem. We need to show that if the winding number of $\gamma$ around every point outside $\Omega$ vanishes, then $\gamma$ integrates every closed holomorphic $1$-form to zero. The strategy uses de Rham cohomology: we show that the winding number forms $dz/(z - z_j)$ — one for each "hole" in $\Omega$ — form a basis for the space of closed $1$-forms modulo exact $1$-forms. Once we know this, any closed $1$-form can be decomposed into a linear combination of winding number forms plus an exact form. The exact form integrates to zero over any closed curve, and each winding number form integrates to $2\pi i \cdot n(\gamma, z_j) = 0$ by hypothesis. So the total integral is zero. Let $\{C_j\}_{j \in J}$ be the bounded connected components of $\mathbb{C} \setminus \Omega$. (The unbounded component does not contribute a generator because it is "at infinity.") Choose representative points $z_j \in C_j$ and define \begin{align*} \omega_j := \frac{1}{2\pi i} \frac{dz}{z - z_j}, \quad j \in J. \end{align*} Why do these form a basis for $H^1_{\mathrm{dR}}(\Omega)$? The topological reason is that $\Omega$, being an open subset of $\mathbb{C}$, deformation retracts onto a graph. Its first homology $H_1(\Omega; \mathbb{Z})$ is free abelian with one generator for each bounded complementary component — geometrically, each "hole" contributes one independent cycle. De Rham's theorem gives $\dim H^1_{\mathrm{dR}}(\Omega) = \operatorname{rank} H_1(\Omega; \mathbb{Z}) = |J|$. To verify linear independence, choose for each $k \in J$ a small circle $\sigma_k$ in $\Omega$ encircling $z_k$ once. The pairing gives \begin{align*} \oint_{\sigma_k} \omega_j = n(\sigma_k, z_j) = \delta_{jk}. \end{align*} This is a non-degenerate pairing (an identity matrix), which forces the $[\omega_j]$ to be linearly independent. The point $z_j$ for $j \neq k$ lies in a connected component of $\mathbb{C} \setminus \Omega$ different from $C_k$. Since $\sigma_k$ is a small circle around $z_k$, all of $C_j$ (for $j \neq k$) lies in the unbounded component of $\mathbb{C} \setminus \operatorname{im}(\sigma_k)$, so by the [Local Constancy of the Winding Number](/theorems/3360), $n(\sigma_k, z_j) = 0$. To verify they span: take any closed holomorphic $1$-form $\omega$ on $\Omega$. Compute its periods $a_k := \oint_{\sigma_k} \omega$. Then $\omega - \sum_j 2\pi i\, a_j\, \omega_j$ has zero period around every $\sigma_k$, hence zero integral around every cycle in $\Omega$ (since the $\sigma_k$ generate $H_1$). A closed $1$-form with zero periods on all cycles is exact (by de Rham's theorem). So $[\omega] = \sum_j 2\pi i\, a_j\, [\omega_j]$ in $H^1_{\mathrm{dR}}(\Omega)$. Now we can decompose any closed holomorphic $1$-form $\omega$ on $\Omega$ as \begin{align*} \omega = \sum_{j \in J} c_j \cdot \frac{dz}{z - z_j} + df \end{align*} for some constants $c_j \in \mathbb{C}$ and some holomorphic function $f: \Omega \to \mathbb{C}$ (the primitive of the exact remainder). Integrating over $\gamma$: \begin{align*} \oint_\gamma \omega = \sum_{j \in J} c_j \oint_\gamma \frac{dz}{z - z_j} + \oint_\gamma df. \end{align*} The second term vanishes because $\gamma$ is a closed curve: $\oint_\gamma df = f(\gamma(b)) - f(\gamma(a)) = 0$ since $\gamma(a) = \gamma(b)$. For the first sum, each integral gives $\oint_\gamma \frac{dz}{z - z_j} = 2\pi i \cdot n(\gamma, z_j)$. By hypothesis, $n(\gamma, z_j) = 0$ for each $j$ (since $z_j \in C_j \subset \mathbb{C} \setminus \Omega$). Therefore \begin{align*} \oint_\gamma \omega = \sum_{j \in J} c_j \cdot 2\pi i \cdot 0 + 0 = 0. \end{align*} Since $\omega$ was arbitrary, $\gamma$ is null-homologous in $\Omega$. [/guided] [/step] [step:Combine both directions to establish the equivalence] The forward direction shows: null-homologous $\implies$ $n(\gamma, z_0) = 0$ for all $z_0 \in \mathbb{C} \setminus \Omega$. The reverse direction shows: $n(\gamma, z_0) = 0$ for all $z_0 \in \mathbb{C} \setminus \Omega$ $\implies$ null-homologous. Together, these establish the claimed equivalence: $\gamma$ is null-homologous in $\Omega$ if and only if $n(\gamma, z_0) = 0$ for all $z_0 \in \mathbb{C} \setminus \Omega$. [/step]