> [!NOTE] Theorem > Assume that $F: \Omega \subset \mathbb{C} \rightarrow \mathbb{C}$ is [[Complex Differentiability|analytic]] ($\Omega$ open) and set $f(z)=\frac{\mathrm{d} F}{\mathrm{~d} z}$, with $f$ continuous. Let $\gamma:[a, b] \rightarrow \Omega$ be a $C^1$ curve. Then $\int_\gamma f \mathrm{~d} z=F(\gamma(b))-F(\gamma(a))$ ###### Proof We have $ \int_\gamma f \mathrm{~d} z=\int_a^b f(\gamma(t)) \gamma^{\prime}(t) \mathrm{d} t=\int_a^b \frac{\mathrm{~d} F}{\mathrm{~d} z}(\gamma(t)) \gamma^{\prime}(t) \mathrm{d} t=\int_a^b \frac{\mathrm{~d}}{\mathrm{~d} t} F(\gamma(t)) \mathrm{d} t=F(\gamma(b))-F(\gamma(a)) $ We remark that there are no assumptions made about $\Omega$ other than it is open. That is, all we need for the result to be true, is that $f$ is analytic in an open neighborhood of the curve. The notion of simply connected (for a domain) will be defined later, but we emphasize that there is no such requirement on $\Omega$ above result to be true.