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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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Identity theorem for holomorphic functions

Statement

If two holomorphic functions on a complex domain agree on a set with an accumulation point in the domain, then they agree everywhere on the domain.

Precisely, let Ω⊆C be a complex domain (A complex domain is a nonempty connected open subset of C), let f,g:Ω→C be holomorphic, and suppose that some a∈Ω is an accumulation point of {z∈Ω:f(z)=g(z)}. Then f=g on Ω. The requirement a∈Ω is essential.

Facts & Assumptions

Given: A complex domain Ω, holomorphic functions f,g:Ω→C, an accumulation point a∈Ω of their agreement set, and the holomorphic difference h:=f−g supplied by Linearity, product, reciprocal, and quotient rules for complex derivatives. A nonempty subset of a connected space that is both open and closed is the whole space (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[L1]

For a holomorphic function h on an open set U, the set of points having a neighbourhood on which h vanishes is both open and closed in U (The locally zero locus of a holomorphic function is clopen).

[L2]

A holomorphic function has finite order m at a exactly when it factors near a as (z−a)mq(z) with q(a)≠0; its order is +∞ exactly when it vanishes on a neighbourhood of a (The order of a zero is the exponent in its local holomorphic factorization).

[L3]

If f:U→C is complex differentiable at a∈U, then f is continuous at a (Complex differentiability at a point implies continuity there).

Proof

technique · direct
1.1givenL2L3

The function h vanishes at points arbitrarily close to a. If it had finite order there, [L2] would give h(z)=(z−a)mq(z) with q(a)≠0, and [L3] would make q nonzero on a smaller neighbourhood; then h would have no zeros there other than possibly a, contrary to accumulation. Hence h has infinite order at a, so [L2] makes it vanish on a neighbourhood of a.

2.1step 1.1L1given

By [L1], the locally zero locus of h is open and closed in Ω; it is nonempty by step 1.1. Since Ω is connected, that locus is all of Ω.

3.1step 2.1∎

Therefore h(z)=0 for every z∈Ω, which means f(z)=g(z) throughout Ω.

Remarks

The accumulation point must belong to the domain. Accumulation only at a boundary point does not force identity, as the companion counterexample shows.

Depends on

Used by

Dependency tree · two levels

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Sources