CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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The contour integral of a constant c is c times the endpoint displacement

Statement

For $c \in C$ and a rectifiable contour $γ : [a, b] \to C$ , $\int_{γ} c d z = c (γ (b) - γ (a)) .$

Facts & Assumptions

Given: A complex constant $c$ and a rectifiable contour $γ$ .

[L1]

Constant and identity functions obey the complex derivative algebra; in particular $(c z)^{'} = c$ (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L2]

Let $F$ be a primitive of a continuous function $f$ on an open set containing the trace of a rectifiable contour $γ : [a, b] \to C$ . If $F^{'} = f$ is continuous, then $\int_{γ} f (z) d z = F (γ (b)) - F (γ (a))$ (The line integral of a continuous function admitting a primitive is that primitive's endpoint increment along every rectifiable path).

Proof

technique · direct

1.1L1

By [L1], $F (z) = c z$ is a primitive of the constant function $c$ .

2.1step 1.1L2algebra∎

The constant function $c$ is continuous on all of $C$ , and $F^{'} = c$ is that same constant, so the hypotheses of [L2] hold on any open set containing the trace. Apply [L2] and simplify $F (γ (b)) - F (γ (a)) = c γ (b) - c γ (a)$ . The cases $c = 0$ and a constant path are included.

Depends on

Dependency tree · 2 levels

Sources

A. Weber, Lecture Notes in Complex Analysis, §1.7