Let $\Gamma$ be a curve in $\mathbb{C}$, parametrised by $\gamma:[a, b] \rightarrow \mathbb{C}$, that is $\gamma([a, b])=\Gamma$. Given $f: \Omega \subset \mathbb{C} \rightarrow \mathbb{C}$ and $\Gamma \subset \Omega$ we have. > [!NOTE] Lemma > If $\gamma^{-}$represents the parametrisation of $\gamma$ in the opposite direction, then > > $ > \int_{\gamma^{-}} f=-\int_\gamma f > $ If a curve $\Gamma$ has attached a sense of direction we will call it a directed curve. In this case we will denote by $-\Gamma$ the same curve swept in the opposite direction. Without the need to specify the parametrisation we can reformulate the above result by $ \int_{\Gamma} f \mathrm{~d} z=-\int_{-\Gamma} f \mathrm{~d} z $ ###### Proof Notice that if $\gamma:[a, b] \rightarrow \mathbb{C}$ parametrises the curve in one direction then $\gamma^{-}$is given by $\gamma^{-}:[a, b] \rightarrow \mathbb{C}$ with $\gamma^{-}(t)=\gamma(a+b-t)$. Therefore $ \begin{aligned} \int_{\gamma^{-}} f & =\int_a^b f\left(\gamma^{-}(t)\right)\left(\gamma^{-}\right)^{\prime}(t) \mathrm{d} t-\int_a^b f(\gamma(a+b-t))\left(-\gamma^{\prime}(a+b-t)\right) \mathrm{d} t \\ & =\int_b^a f(\gamma(s))\left(-\gamma^{\prime}(s)\right)(-1) \mathrm{d} s=-\int_a^b f(\gamma(s)) \gamma^{\prime}(s) \mathrm{d} s=-\int_\gamma f \end{aligned} $