Goursat’s theorem and Cauchy’s theorem

Try

For the triangle with vertices zero, one and i, direct parametrization of exp z would require three separate calculations. Goursat’s theorem gives a more structural answer: the integral is zero because the exponential is holomorphic on a neighborhood of the entire triangle.

The word entire here refers to the filled triangle, including its interior and edges. A function that is holomorphic only near a contour may still have a singularity in the enclosed region. The reciprocal around zero is the permanent test case for that distinction.

Predict

Predict what happens when a triangle is split into four smaller triangles by joining its edge midpoints. Each internal edge is traversed once in each direction when all four boundaries are positively oriented. Those contributions cancel.

If the original integral is nonzero, at least one small triangle must retain a quarter of its modulus. Repeating this choice concentrates the alleged nonzero integral near a single point, where differentiability supplies a strong local approximation.

Experiment

Track the nested triangle chosen at each step.

Step n	Diameter	Perimeter	Guaranteed integral modulus
0	d	L	∣I∣
1	d/2	L/2	At least ∣I∣/4
2	d/4	L/4	At least ∣I∣/16

The two geometric factors both halve. Their product is the crucial factor-of-four decay needed to compete with the lower bound on the integral.

Understand

Let f be holomorphic on an open neighborhood of a closed triangle. Subdivision and the integral triangle inequality select nested closed triangles whose integrals have modulus at least |I|/4ⁿ. Their diameters tend to zero, so they share a single limiting point a. Complex differentiability at a gives f(z)=f(a)+f′(a)(z−a)+(z−a)ε(z), where ε(z) tends to zero.

The constant and linear terms have polynomial primitives, so their closed integrals vanish. For any η>0, sufficiently small selected triangles satisfy |ε(z)|≤η. Their remaining integral has modulus at most η times diameter times perimeter, namely ηdL/4ⁿ. Combining with the lower bound gives |I|≤ηdL. Since η is arbitrary, I=0. Notice that this proof never assumes continuity of f′.

Triangle cancellation constructs primitives on star-shaped domains by integrating from a star center along straight segments. On a simply connected domain, the corresponding global conclusion is Cauchy’s theorem: every closed piecewise smooth contour has zero integral. The extension uses homotopy invariance, subdividing a contraction into patches with local primitives. This is a proof mechanism rather than a full construction of the subdivision; the next lesson separates the topological hypothesis from the local analytic result.

For boundaries of polygonal regions contained with their interiors in the domain, finite triangulation gives direct cancellation. More general piecewise smooth boundaries follow by the equivalent local-primitive contour argument.

Apply

Diagnose the invalid step in “1/z is holomorphic on the unit circle, so its integral is zero.” Holomorphy is a property on open neighborhoods, and even a neighborhood of the circle does not include its filled disk. The origin is missing, so the required contraction or filled-region hypothesis fails.

As a constructive action, draw a square split into two triangles and label the diagonal orientations. Explain algebraically why the diagonal integrals cancel while the outer four edge integrals remain.

Continue

Goursat converts a pointwise derivative definition into a contour identity. The next lesson records which loops are allowed to contract and how holes affect orientation. Optional depth: write the star-shaped primitive construction carefully, checking that the small triangle used in its increment argument stays inside the open domain.