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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
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The weighted argument principle

Statement

Let ΩC be open, let f be meromorphic on Ω, let Γ be admissible for the residue theorem in Ω, suppose f(z)0 for every zΓ, and let g be holomorphic on Ω. Then

12πiΓg(z)f(z)f(z)dz=aZ(f)n(Γ,a)orda(f)g(a)bP(f)n(Γ,b)ordbpole(f)g(b),

and only finitely many terms are nonzero.

Facts & Assumptions

Given: An open set Ω, a meromorphic function f on Ω, an admissible cycle Γ with f0 on Γ, and a holomorphic function g on Ω.

[L1]

The logarithmic derivative has residue m at a zero of order m and residue m at a pole of order m (The logarithmic derivative has residue equal to local order).

[L2]

The unweighted argument principle already shows that only finitely many zeros and poles of f have nonzero index with respect to Γ (The argument principle for an admissible null-homologous cycle).

[L3]

The residue theorem sums the indexed residues of an admissible meromorphic function over Γ (The residue theorem for a null-homologous cycle).

Proof

technique · direct
1.1

Put h(z):=g(z)f(z)/f(z). Away from the zeros and poles of f, the function f/f is holomorphic, so h is holomorphic there as well. At a zero or pole c of f, the function g is holomorphic and therefore admits the expansion g(z)=g(c)+(zc)u(z) near c for some holomorphic u. Multiplying that by the principal-part decomposition from [L1] shows Res(h,c)=g(c)Res ⁣(ff,c).

givenL1algebra
2.1

Step 1.1 and [L1] therefore give Res(h,a)=g(a)orda(f) at each zero a of f, and Res(h,b)=g(b)ordbpole(f) at each pole b of f. By [L2], only finitely many such points have nonzero index with respect to Γ.

step 1.1L1L2
3.1

Applying [L3] to h and substituting the residue values from step 2.1 gives the displayed weighted sum formula.

step 2.1L3

Depends on

Used by

Dependency tree · two levels

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Sources