[proofplan] Since $U$ is simply connected, every closed path in $U$ is null-homotopic. The [Homotopy Invariance of Contour Integrals](/theorems/343) then forces the [integral](/page/Integral) over any closed path to vanish. The existence of an antiderivative follows from the [Antiderivative Existence Characterisation](/theorems/340). [/proofplan] [step:Show every closed-path integral vanishes using null-homotopy] Let $\gamma: [0,1] \to U$ be a closed piecewise $C^1$ path. Since $U$ is simply connected, $\gamma$ is null-homotopic in $U$: there exists a continuous map $H: [0,1]^2 \to U$ with $H(t, 0) = \gamma(t)$, $H(t, 1) = \gamma(0)$ (a constant loop), and $H(0, s) = H(1, s) = \gamma(0)$ for all $s$. By the [Homotopy Invariance of Contour Integrals](/theorems/343), $\gamma$ and the constant loop at $\gamma(0)$ have equal integrals: \begin{align*} \int_\gamma f(z) \, dz = \int_{\text{constant}} f(z) \, dz = 0. \end{align*} The integral over the constant loop vanishes because the loop has zero length. [/step] [step:Construct an antiderivative via the Antiderivative Existence Characterisation] Since $\int_\gamma f(z) \, dz = 0$ for every closed piecewise $C^1$ path $\gamma$ in $U$, the [Antiderivative Existence Characterisation](/theorems/340) provides a holomorphic [function](/page/Function) $F: U \to \mathbb{C}$ with $F' = f$ on $U$. [/step]