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A complex function is holomorphic if and only if it is analytic
Statement
Let be open and let . Then is holomorphic on if and only if it is analytic on in the local power-series sense of Complex analytic functions as locally representable by convergent power series.
Facts & Assumptions
Given: An open set and a function .
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).
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).
Every function analytic on an open subset of is holomorphic there (Every complex analytic function is holomorphic).
Proof
For the holomorphic-to-analytic direction, if is holomorphic and , [L1] gives a positive-radius disc about on which equals its Taylor series, so [L2] makes analytic at and hence on .
For the analytic-to-holomorphic direction, the local power-series hypothesis of [L2] is exactly the hypothesis of [L3], which makes holomorphic on the same open set .
Steps 1.1 and 1.2 prove both implications; when , both pointwise predicates hold vacuously, so the equivalence also includes the empty open set.
Depends on
Used by
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Dependency tree · two levels
17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- B. V. Shabat, Introduction to Complex Analysis, Theorem 2.24 (standard reference, not scraped)
- Matthias Weber, Complex Analysis, §2.2 (standard reference, not scraped)