False statementConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-08-29

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FALSE: Weierstrass factorization is unique

Statement

Weierstrass factorization is unique.

Facts & Assumptions

Given: The constant entire function $1$ .

[F1]

Weierstrass factorization writes an entire function as an exponential factor times a product carrying the zeros (Weierstrass factorization for entire functions).

Refutation

1.1

The function $1$ has no zeros, so [F1] allows the trivial product part and gives the factorization $1 = e^{0} \cdot 1$ .

F1given

2.1

But also $1 = e^{2 π i} \cdot 1$ , and the exponential factors $e^{0}$ and $e^{2 π i}$ come from different entire logarithms. Therefore the factorization is not unique.

step 1.1algebra∎

Depends on

Weierstrass factorization for entire functions

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 The Weierstrass product theorem (standard reference, not scraped)