Proof. Let \(C\) be any contour in \(D\) from \(z_0\) to \(z_1\), parametrised by \(z(t)\) for \(a\leq t\leq b\) with \(z(a)=z_0\) and \(z(b)=z_1\). By the chain rule the composite \(F(z(t))\) has derivative \[\frac {d}{dt}F\big (z(t)\big )=F'\big (z(t)\big )z'(t)=f\big (z(t)\big )z'(t),\] which is exactly the integrand of the contour integral. Hence, by the fundamental theorem of calculus applied to the real and imaginary parts, \[\int _C f(z)\,dz=\int _a^b f\big (z(t)\big )z'(t)\,dt =\Big [F\big (z(t)\big )\Big ]_a^b=F(z_1)-F(z_0).\]

Two consequences are worth naming. The value depends only on the endpoints, so the integral is path independent in \(D\); and taking \(z_1=z_0\) gives \(\int _C f=0\) around any closed contour, which is Cauchy’s theorem in the presence of an antiderivative. □