Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28
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Small perturbations preserve the total local zero multiplicity

Statement

Let f be holomorphic on a neighbourhood of the closed disc D(a,r)‾, and suppose f has no zero on the circle ∣z−a∣=r. Let m be the total multiplicity of the zeros of f in D(a,r). If g is holomorphic on a neighbourhood of D(a,r)‾ and

∣g(z)−f(z)∣<∣f(z)∣(∣z−a∣=r),

then g has exactly m zeros in D(a,r), counted with multiplicity.

In particular, if a is an isolated zero of f of order m and the disc is chosen so that a is the only zero of f in D(a,r)‾, then every such perturbation g has exactly m zeros in D(a,r) counted with multiplicity.

Facts & Assumptions

Given: A holomorphic function f on a neighbourhood of D(a,r)‾, a holomorphic function g on the same neighbourhood, and ∣g−f∣<∣f∣ on ∣z−a∣=r.

[L1]

Rouché's theorem gives equal zero counts inside a closed contour when the strict boundary inequality holds (Rouche's theorem in the classical strict-inequality form).

Proof

technique · direct
1.1given

Let γ(t)=a+reit for 0≤t≤2π. The hypothesis says ∣g−f∣<∣f∣ on γ∗, and f has no zero there.

2.1step 1.1L1

Applying [L1] to the contour γ shows that f and g have the same weighted zero count inside ∣z−a∣<r. Because both are holomorphic, there is no pole term, so that weighted count is exactly the total multiplicity of the interior zeros. Hence g has as many zeros in the disc, counted with multiplicity, as f does.

3.1step 2.1∎

The isolated-zero specialization is the case where that total multiplicity for f is the single local order m at a.

Depends on

Used by

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Sources