# Proof of Goursat's theorem on polygons?

Theorem (Goursat). Let $$U$$ be an open subset of $$\mathbb C$$ and $$f : U \to \mathbb C$$ a complex-differentiable function. Then, for any triangle $$\Delta$$ contained in $$U$$, one has: $$\int_{\partial\Delta}f=0$$ The result can be extended to any polygon contained in $$U$$ by breaking it into sub-triangles.

I've had no issue with the proof on a triangle, however I can't seem to find a rigorous argument to extend the result to any polygon using triangulation.

The exact result I'm trying to prove is that if $$A_1,...,A_n$$ are points in $$U$$ such that the polygon $$A_1\cdots A_n$$ is contained in $$U$$, then: $$\int_{A_1}^{A_2}f+\cdots+\int_{A_{n-1}}^{A_n}f+\int_{A_n}^{A_1}f=0$$

• Actually, from Goursat's theorem, you get easily Cauchy's general version with any loop contained in $U$... Jul 20, 2021 at 22:01
• Notice you traverse every interior edge in both directions... Jul 20, 2021 at 22:10

## 1 Answer

You can use induction. Suppose that it's true for $$n$$-sided polygons. For an $$n+1$$-sided polygon $$A_1, \ldots, A_{n+1}$$, we have the integral

\begin{align*} \int_{A_1}^{A_2} +\cdots+\int_{A_{n}}^{A_{n+1}} + \int_{A_{n+1}}^{A_1} &= \left(\int_{A_1}^{A_2}+\cdots+\int_{A_{n-1}}^{A_n}+\int_{A_{n}}^{A_1}\right)+\left(\int_{A_{n}}^{A_{n+1}} + \int_{A_{n+1}}^{A_1} -\int_{A_{n}}^{A_1} \right)\\ &= \left(\int_{A_1}^{A_2}+\cdots+\int_{A_{n-1}}^{A_n}+\int_{A_{n}}^{A_1}\right)+\left(\int_{A_{n}}^{A_{n+1}} + \int_{A_{n+1}}^{A_1} + \int_{A_{1}}^{A_n} \right). \end{align*}

where $$\int_{a}^{b}$$ is short for $$\int_{a}^{b} f(z) \, dz$$ with the integral over the line segment from $$a$$ to $$b$$. In the first equality we added and then subtracted $$\int_{A_{n}}^{A_1}$$. Then in the last line, both parenthesized summands are $$0$$: the first by the induction hypothesis and the second by the Theorem.

• I have indeed tried this method, the problem I encountered is that the interiors of both triangle $A_n A_{n+1} A_1$ and polygon $A_1\cdots A_n$ need to be contained in $U$ in order to use the induction hypothesis and Goursat's theorem. If your $(n+1)$-gon is not convex for instance, this might not be the case.
– KCJV
Jul 20, 2021 at 22:39
• @KCJV That's a good point I hadn't thought of. One way to fix it for non-convex polygons is to use the Two ears theorem and assume $A_1$ is an "ear". At this point it becomes more a problem of Euclidean geometry rather than complex analysis. Jul 20, 2021 at 23:05