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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
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The Jacobian determinant of a holomorphic map is ∣f′∣2 and is positive exactly where f′≠0

Statement

Let f=u+iv be holomorphic on an open set U⊆C. At every z∈U,

det⁡Jf(z)=∣f′(z)∣2≥0.

The determinant is positive exactly where f′(z)≠0, and it is zero exactly where f′(z)=0; it is never negative.

Facts & Assumptions

Given: A holomorphic map f=u+iv and a point z in its open domain.

[F1]

The Jacobian matrix is the matrix of the first partial derivatives (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

[L2]

Complex modulus satisfies ∣a+ib∣2=a2+b2, and ∣w∣=0 if and only if w=0 (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct
1.1

By [L1] and [F1], det⁡Jf(z)=a2+b2=∣f′(z)∣2.

givenL1F1L2algebra
2.1

The sum a2+b2 is nonnegative and, by [L2], is zero exactly when f′(z)=0; otherwise it is positive.

step 1.1L2algebra∎

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