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Goursat's theorem for rectangles: a holomorphic function integrates to zero around every rectangle contained in its domain

Statement

Let $U \subseteq C$ be open, let $f : U \to C$ be holomorphic, and fix $a \in C$ and real numbers $w, h > 0$ . Suppose the closed rectangle

$R = {a + x + i y : 0 \leq x \leq w, 0 \leq y \leq h}$

is contained in $U$ . Put $b = a + w$ , $c = a + w + i h$ , and $d = a + i h$ . Its positively oriented boundary is the closed rectifiable contour

$\partial R = ℓ_{a b} * ℓ_{b c} * ℓ_{c d} * ℓ_{d a}$

using the directed segments and concatenation of Filled complex triangles, their oriented three-edge boundary contours, diameter, and perimeter and Rectifiable complex contours, reversal, concatenation, closedness, and orientation. Then

$\int_{\partial R} f (z) d z = 0.$

Facts & Assumptions

Given: The rectangle $R \subseteq U$ , its ordered vertices $a, b, c, d$ , its positively oriented boundary as displayed, and a holomorphic $f : U \to C$ .

[L1]

A holomorphic function integrates to zero around the oriented boundary of every filled triangle contained in its open domain (Goursat's triangle theorem: a holomorphic function integrates to zero around every triangle contained in its domain).

[L2]

Reversal negates a contour integral and concatenation adds contour integrals (Complex line integrals change sign under reversal and add under concatenation).

Proof

technique · direct

1.1given

The diagonal from $a$ to $c$ splits $R$ into the filled triangles $Δ [a, b, c]$ and $Δ [a, c, d]$ , both contained in $U$ .

2.1step 1.1L1

By [L1], the integrals over the positively oriented boundaries $a \to b \to c \to a$ and $a \to c \to d \to a$ are both zero.

3.1step 2.1L2

Adding those identities, the diagonal $c \to a$ in the first boundary cancels the diagonal $a \to c$ in the second by [L2].

4.1step 3.1L2∎

The surviving directed sides are $a \to b \to c \to d \to a$ , exactly the displayed positive boundary $\partial R$ , so its integral is zero.

Depends on

Used by

Admissible cycle around a compact plane set Lemma

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

E. Stein and R. Shakarchi, Complex Analysis, Ch. 2, Corollary 1.2 (standard reference, not scraped)