[proofplan] We construct the holomorphic logarithm as a path integral $\lambda(z) = \lambda_0 + \int_\gamma \zeta^{-1} \, d\zeta$ from a fixed basepoint $z_0$. Simple connectedness of $D$ and [Cauchy's theorem for simply connected domains](/theorems/344) guarantee path-independence. Holomorphicity of $\lambda$ follows by differentiating the integral. To verify $e^{\lambda(z)} = z$, we show $z \cdot e^{-\lambda(z)}$ has zero derivative on the connected domain $D$ and equals $1$ at the basepoint, hence is identically $1$. [/proofplan] [step:Define $\lambda(z)$ as a path integral and verify well-definedness via Cauchy's theorem] Fix a basepoint $z_0 \in D$ and choose $\lambda_0 \in \mathbb{C}$ with $e^{\lambda_0} = z_0$. For $z \in D$, let $\gamma$ be any piecewise $C^1$ path in $D$ from $z_0$ to $z$, and define \begin{align*} \lambda(z) = \lambda_0 + \int_\gamma \frac{1}{\zeta} \, d\zeta. \end{align*} If $\gamma_1$ and $\gamma_2$ are two paths from $z_0$ to $z$, then $\gamma_1 * \overline{\gamma_2}$ is a closed loop in $D$. Since $D$ is simply connected, this loop is null-homotopic in $D$. The [function](/page/Function) $\zeta \mapsto 1/\zeta$ is holomorphic on $D \subseteq \mathbb{C}^*$, so by [Cauchy's theorem for simply connected domains](/theorems/344): \begin{align*} \int_{\gamma_1} \frac{d\zeta}{\zeta} - \int_{\gamma_2} \frac{d\zeta}{\zeta} = \oint_{\gamma_1 * \overline{\gamma_2}} \frac{d\zeta}{\zeta} = 0. \end{align*} Therefore $\lambda(z)$ is independent of the choice of path. [guided] Why is simple connectedness needed? The function $\zeta \mapsto 1/\zeta$ is holomorphic on $D \subseteq \mathbb{C}^*$, but Cauchy's theorem requires a topological condition on the domain to guarantee vanishing of closed-path integrals. Specifically, we need every closed loop in $D$ to be null-homotopic. On a non-simply-connected domain like $\mathbb{C}^*$ itself, the loop $\gamma_1 * \overline{\gamma_2}$ might wind around $0$, and the integral $\oint_{\gamma_1 * \overline{\gamma_2}} \zeta^{-1} \, d\zeta = 2\pi i \cdot n(\gamma, 0)$, where $n(\gamma, 0)$ is the [winding number](/theorems/351). If the two paths $\gamma_1, \gamma_2$ wind differently around $0$, this integral is non-zero and the definition of $\lambda(z)$ depends on the choice of path. On a simply connected subdomain of $\mathbb{C}^*$, every closed loop is contractible (null-homotopic), so by [Cauchy's theorem for simply connected domains](/theorems/344) the integral around any closed loop vanishes. In particular, $\oint_{\gamma_1 * \overline{\gamma_2}} \zeta^{-1} \, d\zeta = 0$, and the winding number around $0$ is forced to be $0$. This is the topological obstruction to defining a single-valued logarithm: on $\mathbb{C}^*$, the integral $\int \zeta^{-1} \, d\zeta$ picks up a $2\pi i$ ambiguity for each winding around the origin. Different paths from $z_0$ to $z$ can yield values of $\lambda(z)$ differing by integer multiples of $2\pi i$, making the logarithm multi-valued. Simple connectedness eliminates this ambiguity by ensuring all paths are homotopic, so the integral is path-independent. [/guided] [/step] [step:Show $\lambda$ is holomorphic with $\lambda'(z) = 1/z$] For $z \in D$, choose $\varepsilon > 0$ with $B(z, \varepsilon) \subseteq D$. For $|h| < \varepsilon$, take the path from $z_0$ to $z$ followed by the straight segment from $z$ to $z + h$. Then \begin{align*} \lambda(z + h) - \lambda(z) = \int_z^{z+h} \frac{d\zeta}{\zeta} = \int_0^1 \frac{h}{z + th} \, d\mathcal{L}^1(t). \end{align*} The integrand $h/(z + th)$ converges uniformly on $[0,1]$ to $h/z$ as $h \to 0$ (since $|z + th| \geq |z| - |h| \geq |z|/2$ for $|h|$ small enough). Therefore \begin{align*} \lambda(z + h) - \lambda(z) = \frac{h}{z} + o(|h|), \end{align*} giving $\lambda'(z) = 1/z$. [/step] [step:Verify $e^{\lambda(z)} = z$ using constancy on a connected domain] Define $g: D \to \mathbb{C}$ by $g(z) = z \cdot e^{-\lambda(z)}$. By the product rule and $\lambda'(z) = 1/z$: \begin{align*} g'(z) = e^{-\lambda(z)} + z \cdot (-\lambda'(z)) e^{-\lambda(z)} = e^{-\lambda(z)} - z \cdot \frac{1}{z} e^{-\lambda(z)} = 0. \end{align*} Since $D$ is connected and $g'(z) = 0$ for all $z \in D$, the [Zero Derivative Implies Constancy](/theorems/329) theorem gives that $g$ is constant on $D$. At the basepoint: $g(z_0) = z_0 e^{-\lambda_0} = z_0 / z_0 = 1$. Therefore $z \cdot e^{-\lambda(z)} = 1$ for all $z \in D$, i.e., $e^{\lambda(z)} = z$. [/step] [step:Characterise the non-uniqueness of the branch] If $\lambda_1$ and $\lambda_2$ are two holomorphic logarithms on $D$, then $e^{\lambda_1(z)} = z = e^{\lambda_2(z)}$, so $e^{\lambda_1(z) - \lambda_2(z)} = 1$ for all $z \in D$. The [function](/page/Function) $\lambda_1 - \lambda_2$ is holomorphic on $D$ with values in $2\pi i \mathbb{Z}$ (since $e^w = 1$ iff $w \in 2\pi i \mathbb{Z}$). Since $D$ is connected and $\lambda_1 - \lambda_2$ is [continuous](/page/Continuity), its image in the discrete set $2\pi i \mathbb{Z}$ must be a single point. Therefore $\lambda_1 - \lambda_2$ is a constant in $2\pi i \mathbb{Z}$. [/step]