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Cauchy's integral formula for a null-homologous cycle

Statement

Let ΩC be open, let f:ΩC be holomorphic, and let Γ be a complex chain which is a cycle, with trace in Ω and null-homologous in Ω. Then for every zΩΓ

n(Γ,z)f(z)=12πiΓf(ζ)ζzdζ.

No connectedness of Ω is assumed.

Facts & Assumptions

Given: An open Ω, a holomorphic f:ΩC, and a cycle Γ with ΓΩ which is null-homologous in Ω.

[L1]

With g the filled difference quotient of f, the function h equal to (2πi)1Γg(ζ,z)dζ on Ω and to (2πi)1Γf(ζ)(ζz)1dζ on Ω0={zΓ:n(Γ,z)=0} is a well-defined entire function; it is bounded, and for every ε>0 there is R>0 with h(z)<ε whenever z>R (Dixon's glued function is entire and vanishes at infinity).

[L2]

Every bounded entire function is constant (Liouville's theorem: every bounded entire function is constant).

[L3]

If f is continuous on the trace of a complex chain, then Γfdz=k<r,mk0mkγkfdz; and for zΓ one has n(Γ,z)=(2πi)1Γdζ/(ζz) (Integration over a complex chain and the index of a chain). A chain is a finite list of integer-weighted contours with trace the union of the γk having mk0 (Complex chains, their traces, and cycles).

[L4]

A cycle Γ with trace in Ω is null-homologous in Ω when n(Γ,p)=0 for every pCΩ (Null-homologous cycles and homologous cycles in an open set).

[L5]

Complex line integrals are linear in the integrand (Complex line integrals are linear in the integrand); finite sums in the additive commutative monoid of C are additive, and complex-field distributivity permits scaling term by term (A finite sum in a commutative monoid indexed by an arbitrary finite set, C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (abi)/(a2+b2)).

[L6]

The filled difference quotient g of a holomorphic f on Ω equals (f(ζ)f(z))/(ζz) off the diagonal and f(z) on it (The filled difference quotient of a holomorphic function is jointly continuous).

[L8]

Proof

technique · direct
1.1

By [L1] the glued function h is entire and bounded, so [L2] makes it a constant c.

givenL1L2L7
1.2

By [L1], for every ε>0 there is R>0 with h(z)<ε for z>R; such z exist, so c<ε for every ε>0 and therefore c=0.

givenL1
1.3

Let zΩΓ. Since f is holomorphic on Ω, [L8] makes it continuous on Ω, hence on the trace of Γ. Then ζz for every ζΓ, so [L6] gives g(ζ,z)=(f(ζ)f(z))/(ζz) on the trace, and [L3] with [L5] splits the defining integral into h(z)=(2πi)1Γf(ζ)(ζz)1dζf(z)n(Γ,z).

givenL3L5L6L8
2.1

Steps 1.1, 1.2 and 1.3 give 0=(2πi)1Γf(ζ)(ζz)1dζn(Γ,z)f(z), which is the stated formula; nothing in the argument used connectedness of Ω, and the hypothesis that Γ is null-homologous entered only through [L1] and [L4].

step 1.1step 1.2step 1.3L1L4

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