Show that if \(f\) has a primitive \(F\) on \(\Omega\) then \(\displaystyle\int_\gamma f = 0\) for every closed curve \(\gamma \subseteq \Omega\).
Solution
Let \(F\) be a primitive of \(f\), so \(\dd{}{z}F = f\). Then \[ \int_\gamma f(z) \dz = F(\gamma(1)) - F(\gamma(0)) = F(p) - F(p) = 0 .\] More explicitly, let \(z(t): [a, b]\to \CC\) be any parameterization of \(\gamma\), then \[ \int_\gamma f(z) \dz &= \int_a^b f(z(t)) z'(t)\dt \\ &= \int_a^b F'(z(t))z'(t) \dt \\ &= \int_a^b \tilde F'(t)\dt && \text{ where } \tilde F(t) \da F(z(t)) \text{ by the chain rule} \\ &= F(z(b)) - F(z(a)) && \text{ by FTC} \\ &= 0 ,\] since \(z(b) = z(a)\) for a closed curve.