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Cauchy's integral theorem

Statement

Let U⊂CU \subset \mathbb{C} be a simply connected open domain and let f:U→Cf : U \to \mathbb{C} be a holomorphic function. Then for every piecewise smooth closed contour γ\gamma in UU, the contour integral vanishes: ∮γf(z) dz=0\oint_{\gamma} f(z)\,dz = 0.

Why is it true?

In multivariable calculus, a line integral around a closed loop is zero when the vector field is conservative — meaning it has no local curl or 'swirl' inside the loop. Complex differentiability is an extremely rigid condition: the Cauchy–Riemann equations force both the real and imaginary parts of f(z) dzf(z)\,dz to be curl-free simultaneously. As long as the region has no holes where a singularity could hide and create circulation, integrating f(z)f(z) along any closed path cancels out completely, and integrals between two points depend only on the endpoints.

Proof sketch

Cauchy's original 1825 approach assumes f′f' is continuous, writes f(z) dz=(u dx−v dy)+i(v dx+u dy)f(z)\,dz = (u\,dx - v\,dy) + i(v\,dx + u\,dy), and applies Green's theorem so the double integrals vanish by the Cauchy–Riemann equations ∂u/∂x=∂v/∂y\partial u/\partial x = \partial v/\partial y and ∂u/∂y=−∂v/∂x\partial u/\partial y = -\partial v/\partial x. In 1900, Édouard Goursat removed the continuity assumption on f′f': one first proves the integral around any triangle T⊂UT \subset U vanishes by bisecting TT into four sub-triangles, picking the one with the largest integral modulus, and using the local linear approximation f(z)=f(z0)+f′(z0)(z−z0)+o(∣z−z0∣)f(z) = f(z_0) + f'(z_0)(z - z_0) + o(|z - z_0|) at the intersection point z0z_0 to show the integral is bounded by an arbitrarily small multiple of the perimeter squared. From triangles one builds a primitive on star-shaped domains and extends to general simply connected domains.

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Step-by-step proofs

No step-by-step proof yet for this theorem.

References

  1. Lars V. Ahlfors (1979). Complex Analysis
  2. Reinhold Remmert (1991). Theory of Complex Functions