1
$\begingroup$

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: $$\int_{A_1}^{A_2}f+\cdots+\int_{A_{n-1}}^{A_n}f+\int_{A_n}^{A_1}f=0$$

$\endgroup$
2
  • 2
    $\begingroup$ Actually, from Goursat's theorem, you get easily Cauchy's general version with any loop contained in $U$... $\endgroup$ Jul 20, 2021 at 22:01
  • 4
    $\begingroup$ Notice you traverse every interior edge in both directions... $\endgroup$ Jul 20, 2021 at 22:10

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – KCJV
    Jul 20, 2021 at 22:39
  • 1
    $\begingroup$ @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. $\endgroup$ Jul 20, 2021 at 23:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .