Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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A real-differentiable complex map is orientation-preserving conformal at a point exactly when it is complex differentiable there with nonzero derivative

Statement

Let f:UC be real totally differentiable at aU. Then f is orientation-preserving conformal at a if and only if it is complex differentiable at a and f(a)0.

Facts & Assumptions

Given: An open UC, a point aU, and a map f:UC real totally differentiable at a.

[F1]

Orientation-preserving conformality at a means that Df(a) is an orientation-preserving similarity (Orientation-preserving conformality for a real-differentiable complex map at a point).

[L2]

The orientation-preserving similarities of the plane are exactly the maps hξh with ξ0 (Plane similarities are complex or conjugate-complex multiplications; the orientation-preserving ones are exactly the nonzero complex multiplications).

Proof

technique · direct
1.1

If f is complex differentiable at a with f(a)0, [L1] makes Df(a) multiplication by f(a), and [L2] makes this an orientation-preserving similarity. Hence f is conformal at a by [F1].

givenL1L2F1
1.2

Conversely, if f is orientation-preserving conformal at a, [F1] and [L2] give Df(a)h=ξh for some ξ0. The reverse direction of [L1] makes f complex differentiable with f(a)=ξ0.

givenF1L2L1
2.1

Steps 1.1 and 1.2 prove both directions of the equivalence.

step 1.1step 1.2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 38 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources