TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

On positive reals, the principal branch power agrees with the published real power

Statement

For $x > 0$ and real $α \in R$ , the principal branch power $x_{Log}^{} : = \exp (α Log x)$ of Complex powers defined from a holomorphic logarithm branch equals the published real power $x^{α}$ of Real powers for positive bases, with the zero-base positive-exponent convention:

$x_{Log}^{} = x^{α} (x > 0, α \in R) .$

Facts & Assumptions

Given: A positive real $x > 0$ and a real exponent $α \in R$ .

[F1]

On the slit plane, the principal branch power is $z_{Log}^{} : = \exp (α Log z)$ , where $Log$ is the holomorphic principal logarithm branch (Complex powers defined from a holomorphic logarithm branch).

[F2]

For $a > 0$ and $x \in R$ , the real power is $a^{x} : = \exp (x \log a)$ (Real powers for positive bases, with the zero-base positive-exponent convention).

[F3]

For real $x$ , the complex value $\exp (x + 0 i)$ equals the published real exponential $e^{x}$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

Proof

technique · direct

1.1

$Log x = \log x + i 0$ , the real logarithm embedded in $C$ : the polar form of $x > 0$ uses angle $0$ .

F1given

2.1

Steps 1.1, [F1], [F3], and [F2] give $x_{Log}^{} = \exp ((α \log x) + i 0) = e^{α \log x} = x^{α}$ .

F1F2F3step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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

Lars V. Ahlfors, Complex Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)