[proofplan] The [Chain Rule](/theorems/323) gives $\frac{d}{dt}F(\gamma(t)) = F'(\gamma(t)) \cdot \gamma'(t) = f(\gamma(t)) \cdot \gamma'(t)$, which is the integrand of the contour [integral](/page/Integral). The real [Fundamental Theorem of Calculus](/theorems/632) (applied to real and imaginary parts separately) then evaluates $\int_a^b \frac{d}{dt}F(\gamma(t)) \, d\mathcal{L}^1(t) = F(\gamma(b)) - F(\gamma(a))$. [/proofplan] [step:Differentiate the composition $F \circ \gamma$ and apply the Fundamental Theorem of Calculus] By the [Chain Rule](/theorems/323), on each smooth piece of $\gamma$, the [function](/page/Function) $t \mapsto F(\gamma(t))$ is [differentiable](/page/Derivative) with \begin{align*} \frac{d}{dt}F(\gamma(t)) = F'(\gamma(t)) \cdot \gamma'(t) = f(\gamma(t)) \cdot \gamma'(t), \end{align*} since $F' = f$ by hypothesis. Applying the real [Fundamental Theorem of Calculus](/theorems/632) to the real and imaginary parts of $F \circ \gamma$ on each smooth piece, and summing over pieces: \begin{align*} \int_\gamma f(z) \, dz = \int_a^b f(\gamma(t)) \gamma'(t) \, d\mathcal{L}^1(t) = \int_a^b \frac{d}{dt} F(\gamma(t)) \, d\mathcal{L}^1(t) = F(\gamma(b)) - F(\gamma(a)). \end{align*} If $\gamma$ is closed, then $\gamma(b) = \gamma(a)$, so $\int_\gamma f(z) \, dz = 0$. [/step]