Goursat's triangle argument
/ Goursat: goor-SAH /
This is the clever engine inside the Cauchy-Goursat theorem — the concrete argument that proves a holomorphic function's integral around a triangle is zero using nothing but differentiability. It is worth meeting in detail because it shows, in miniature, how a single derivative at each point controls a global integral.
Here is the method in plain steps. Start with a triangle T and call the size of its boundary integral I (we want to show I = 0). Connect the midpoints of T's three sides to cut it into four congruent sub-triangles. The four sub-boundary integrals add up to the integral over T, because the new interior edges are each traversed twice in opposite directions and cancel. By the triangle inequality, at least one of the four sub-integrals has size at least |I|/4. Pick that sub-triangle and repeat, getting a nested sequence of ever-smaller triangles, the n-th one carrying an integral of size at least |I|/4^n while having a quarter the diameter each time. The triangles shrink to a single point p. Near p, differentiability says f(z) = f(p) + f'(p)(z - p) + (a remainder that is small compared to |z - p|). The constant and linear parts integrate to exactly zero around any closed loop (they have primitives), so only the small remainder contributes; the ML inequality bounds the n-th integral by (something tiny) times (perimeter), which beats |I|/4^n unless |I| = 0. Hence I = 0.
The beauty is that no continuity of f' was ever used — only the defining limit of the derivative at the single limit point p. From triangles the result spreads to polygons (cut into triangles) and then to general contours (approximate by polygons), giving the full theorem. This argument is the reason complex analysis can claim its central miracle on the mildest of hypotheses.
If the n-th nested triangle has diameter d_n (a quarter of the previous) and perimeter L_n (also a quarter), then near p the remainder is at most epsilon times d_n, so its integral is at most epsilon times d_n times L_n, which is (epsilon times d_0 times L_0)/4^n. This must be at least |I|/4^n, forcing |I| at most epsilon times d_0 times L_0 for every epsilon, hence |I| = 0.
How the 4^n bookkeeping cancels against the 4^n shrinkage, leaving only zero.
The argument proves the loop integral is zero only because the constant and linear terms of f's local expansion already have primitives; it is the higher-order smallness of the remainder, not any magic, that does the work.