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A closed contour path-homotopic to a constant loop has zero integral against every holomorphic function

Statement

Let $Ω \subseteq C$ be open, let $f : Ω \to C$ be holomorphic, and let $γ : [0, 1] \to Ω$ be a closed rectifiable contour. If $γ$ is path-homotopic relative to the endpoints to a constant loop in $Ω$ , then

$\int_{γ} f (z) d z = 0.$

Facts & Assumptions

Given: An open set $Ω$ , a holomorphic function $f : Ω \to C$ , a closed rectifiable contour $γ : [0, 1] \to Ω$ , and a constant loop $c : [0, 1] \to Ω$ to which $γ$ is path-homotopic relative to the endpoints.

[L1]

A path homotopy relative to the endpoints keeps the two endpoint values fixed throughout the homotopy (Homotopies of continuous maps, homotopies relative to a subspace, and path homotopies relative to the endpoints).

[L2]

Endpoint-fixed homotopic rectifiable paths have equal holomorphic line integrals (Endpoint-fixed homotopic paths have equal holomorphic line integrals).

[L3]

The contour integral of a constant integrand over any contour is that constant times the endpoint displacement (The contour integral of a constant c is c times the endpoint displacement).

[L4]

A closed contour has the same initial and terminal point, and constant paths are legitimate contours (Rectifiable complex contours, reversal, concatenation, closedness, and orientation).

Proof

technique · direct

1.1

The Given and [L1] place $γ$ and the constant loop $c$ under the hypotheses of [L2]. Therefore $\int_{γ} f (z) d z = \int_{c} f (z) d z .$

givenL1L2

2.1

Because $c$ is constant, one has $f (c (t)) = f (c (0))$ for every $t$ . Since [L4] makes $c$ a closed contour, [L3] gives $\int_{c} f (z) d z = \int_{c} f (c (0)) d z = f (c (0)) (c (1) - c (0)) = 0.$ Combining this with step 1.1 proves $\int_{γ} f (z) d z = 0$ .

step 1.1L3L4algebra∎

Depends on

Used by

A plane domain with trivial fundamental group is homologically simply connected Lemma

Dependency tree · two levels

18 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. 3, §5 (standard reference, not scraped)