Goursat's theorem

Statement

Let $U \subseteq \mathbb{C}$ be an open set and let $f : U \to \mathbb{C}$ be holomorphic on $U$. Let $Q = \{x + iy : a_1 \le x \le b_1, \ a_2 \le y \le b_2\}$ be a closed rectangle with $Q \subseteq U$, where $a_1 < b_1$ and $a_2 < b_2$ are real numbers. Let $\partial Q$ be the boundary of $Q$, traveled once counterclockwise. Then the contour integral of $f$ along $\partial Q$ is $$\oint_{\partial Q} f(z) \, dz = 0.$$

Remarks

The theorem does not assume that $f'$ is continuous. This is the difference with the older proof by Green's theorem. The same statement holds for each closed triangle $T \subseteq U$ and its boundary. Proof idea: divide $Q$ into four equal rectangles and keep one where the integral is largest. Repeat, and use the complex derivative at the limit point. Goursat's theorem is the first step to primitives on a disc and to Cauchy's integral theorem.

Explanations

L1 human

If $f$ is holomorphic on a rectangle and its inside, then the integral of $f$ around the edge of the rectangle is zero. The surprise is that this needs only the existence of the complex derivative, and nothing about its continuity.