# Statements > [!NOTE] Theorem (Formulation on simply connected regions) \[Analysis 3\] > Let $f: \Omega \rightarrow \mathbb{C}$ be a [[Complex Differentiability|holomorphic]] function, with $\Omega$ an open, [[Simple connectedness|simply connected]] domain. Let $\gamma$ be a $C^1$ closed curve in $\Omega$. Then $\int_\gamma f(z) \mathrm{d} z=0.$ # Proofs ###### Proof \[Analysis 3\] The proof presented here assumes that the curve a simple, regular curve and that $f^{\prime}$ is continuous. If the domain is simply connected, i.e. the region inside the curve does not have any holes, and $f$ is analytic in it. By [[Green's theorem]], $ \begin{aligned} & \int_\gamma f \mathrm{~d} z=\operatorname{circulation}(\underline{f})+\mathrm{i} \text { flux }(\underline{f}) \\ & =\iint_{\Omega} \operatorname{curl} \underline{f} \mathrm{~d} x \mathrm{~d} y+\mathrm{i} \iint_{\Omega} \operatorname{div} \underline{f} \mathrm{~d} x \mathrm{~d} y \end{aligned} $ We claim that both terms are actually $0$, because $\operatorname{curl} \underline{f}=\operatorname{div} \underline{f}=0$. Since $\underline{f}=(u,-v)$ we have $ \operatorname{div} \underline{f}=u_x-v_y \quad \operatorname{curl} \underline{f}=-v_x-u_y $ but since $f=u+\mathrm{i} v$ is analytic it satisfies the [[Cauchy-Riemann Equations]], $u_x=v_y \quad v_x=-u_y$ which imply the result. # Applications # References