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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
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A contour formula for a locally single-valued holomorphic inverse

Statement

Let ΩC be open, let f be holomorphic on Ω, let Γ be a closed complex contour null-homologous in Ω, and let wC satisfy f(z)w for every zΓ. Suppose f(z)=w has exactly one solution a inside Γ, that solution is simple, and n(Γ,a)=1. Then

a=12πiΓζf(ζ)f(ζ)wdζ.

Thus, on a contour enclosing exactly one simple preimage branch, the inverse value is recovered by a contour integral.

Facts & Assumptions

Given: A holomorphic function f on an open set Ω, a closed null-homologous contour Γ, and a value w satisfying the hypotheses of the statement.

[L1]

The weighted argument principle multiplies each zero contribution by the value of the holomorphic test function there (The weighted argument principle).

[L2]

The preimage-count corollary identifies the zeros of fw inside Γ (The argument principle counts preimages of a target value).

Proof

technique · direct
1.1

Because f is holomorphic, the meromorphic function fw has no poles. The hypotheses say that its only zero inside Γ is the simple zero a.

given
2.1

Apply [L1] to the meromorphic function fw and the holomorphic test function g(ζ)=ζ. By step 1.1, there is only one zero contribution, its multiplicity is 1, and the additional hypothesis n(Γ,a)=1 makes that contribution exactly a. The left-hand side is exactly the displayed contour integral.

step 1.1L1
3.1

Therefore the contour integral equals a. The role of [L2] is to identify the unique enclosed zero as the unique preimage of w.

step 1.1step 2.1L2

Depends on

Used by

Dependency tree · two levels

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Sources