This allows all closed line integrals to be transformed to circle integrals. > [!NOTE] Corollary > Let $\Omega \subset \mathbb{C}$ be a region bounded by two simple curves $\gamma_1$ (the exterior curve) and $\gamma_2$ (the interior). Assume they are oriented positively, and let $f$ be an analytic function in $\Omega \cup \gamma_1 \cup \gamma_2$. Then $\int_{\gamma_1} f \mathrm{~d} z+\int_{\gamma_2} f \mathrm{~d} z=0$ > > **NOTE**: If we denote by $\gamma_2^{-}$the anti-clockwise parametrization of $\gamma_2$, then the result can be rephrased as $\int_{\gamma_1} f \mathrm{~d} z=\int_{\gamma_2^{-}} f \mathrm{~d} z$that is the integral is the same along both curves when both are parametrised counter-clockwise since $\int_{\gamma_2} f \mathrm{~d} z=-\int_{\gamma_2^{-}} f \mathrm{~d} z.$ ###### Proof The proof is based on creating two new contours of integration, the boundaries of two simply connected regions where $f$ is analytic so that we can apply [[Cauchy-Goursat Theorem]]. ![[Screenshot 2025-04-09 at 17.33.17.png|300]] To achieve this we add two new curves to the previous picture, now in yellow in Figure 7.3. They join the points $A$ (in $\gamma_{1}$) with $D$ (in $\gamma_2$) and the points $B$ (in $\gamma_{1}$) with $C$ (in $\gamma_2$). The two curves we want to consider are denoted by $\rho$ and $\eta$. Each one of them is piecewise $C^1$ and formed by four sections. Each one of these curves is oriented positively with respect to the region they enclose, that is, they are both oriented counter-clockwise. By Cauchy's Theorem $\begin{align} & \int_\rho f \mathrm{~d} z=\int_{\rho_1} f \mathrm{~d} z+\int_{\rho_2} f \mathrm{~d} z+\int_{\rho_3} f \mathrm{~d} z+\int_{\rho_4} f \mathrm{~d} z=0\tag{1} \\ & \int_\eta f \mathrm{~d} z=\int_{\eta_1} f \mathrm{~d} z+\int_{\eta_2} f \mathrm{~d} z+\int_{\eta_3} f \mathrm{~d} z+\int_{\eta_4} f \mathrm{~d} z=0\tag{2} \end{align}$ We observe that $\eta_1$ and $\rho_4$ correspond to the same curve but with parametrisations in opposite directions. Similarly for $\eta_3$ and $\rho_2$. Therefore $ \int_{\eta_1} f \mathrm{~d} z+\int_{\rho_4} f \mathrm{~d} z=0 \quad \int_{\eta_3} f \mathrm{~d} z+\int_{\rho_2} f \mathrm{~d} z=0 . $ Adding $(1)$ and $(2)$ and using the above identities we find $\int_{\rho_1} f \mathrm{~d} z+\int_{\rho_3} f \mathrm{~d} z+\int_{\eta_2} f \mathrm{~d} z+\int_{\eta_4} f \mathrm{~d} z=0$ Also notice that $\rho_1$ and $\eta_4$ together build $\gamma_1$, while $\rho_3$ and $\eta_2$ build $\gamma_2$. Therefore, the above equality can be rewritten as $\int_{\gamma_1} f \mathrm{~d} z+\int_{\gamma_2} f \mathrm{~d} z=0$