the route to Cauchy's theorem via Green's theorem
/ Green; Cauchy: koh-SHEE /
Before Goursat's lean triangle argument, there is an older and very intuitive way to see why Cauchy's theorem is true — Cauchy's own route, through Green's theorem from multivariable calculus. It is worth knowing because it makes the Cauchy-Riemann equations visibly responsible for the integral vanishing, connecting the complex theorem to a real-plane theorem you may already have met.
Here is the chain of reasoning. Write f = u + iv and z = x + iy, so dz = dx + i dy. Multiplying out, f dz = (u dx - v dy) + i (v dx + u dy), turning the single complex contour integral into two real line integrals around the closed curve. Green's theorem converts each real line integral around a closed curve into a double integral of partial derivatives over the enclosed region. When you carry this out, the two resulting area integrands are (-v_x - u_y) and (u_x - v_y). But the Cauchy-Riemann equations say u_x = v_y and u_y = -v_x — which make both integrands identically zero. So both double integrals vanish, and the contour integral is zero.
This is genuinely illuminating: Cauchy's theorem is, in this light, the Cauchy-Riemann equations integrated over a region. The catch, and the reason Goursat's proof superseded it, is that Green's theorem in this form requires the partial derivatives u_x, u_y, v_x, v_y to be continuous — equivalently, it assumes f' is continuous. Goursat's quadrisection argument later removed that assumption, proving the same theorem from bare complex-differentiability. So the Green's-theorem route is the honest, motivating proof under a slightly stronger hypothesis; Goursat is the same conclusion made airtight.
For f(z) = z = x + iy, here u = x, v = y, so u_x = 1, v_y = 1, u_y = 0, v_x = 0. The Green's-theorem area integrands (-v_x - u_y) and (u_x - v_y) become (0 - 0) and (1 - 1), both zero, confirming the integral of z around any closed curve is 0 — consistent with z having primitive z^2/2.
Green's theorem plus Cauchy-Riemann makes both area integrands vanish.
This route quietly assumes f' is continuous (so the partials are continuous and Green's theorem applies); that extra hypothesis, which Goursat showed is unnecessary, is the only thing separating it from the fully general proof.