Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21
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A complex function is holomorphic if and only if it is analytic

Statement

Let UC be open and let f:UC. Then f is holomorphic on U if and only if it is analytic on U in the local power-series sense of Complex analytic functions as locally representable by convergent power series.

Facts & Assumptions

Given: An open set UC and a function f:UC.

[L1]

Every holomorphic function equals its Taylor series throughout the largest centred open disc contained in its domain (A holomorphic function equals its Taylor series throughout the largest centred disc in its domain).

[L2]

A function is analytic on an open set exactly when every point has a contained open disc on which the function equals a convergent complex power series centred at that point (Complex analytic functions as locally representable by convergent power series).

[L3]

Every function analytic on an open subset of C is holomorphic there (Every complex analytic function is holomorphic).

Proof

technique · direct
1.1

For the holomorphic-to-analytic direction, if f is holomorphic and aU, [L1] gives a positive-radius disc about a on which f equals its Taylor series, so [L2] makes f analytic at a and hence on U.

L1L2
1.2

For the analytic-to-holomorphic direction, the local power-series hypothesis of [L2] is exactly the hypothesis of [L3], which makes f holomorphic on the same open set U.

L2L3
2.1

Steps 1.1 and 1.2 prove both implications; when U=, both pointwise predicates hold vacuously, so the equivalence also includes the empty open set.

step 1.1step 1.2

Depends on

Used by

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Dependency tree · two levels

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Sources