Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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The unit circle traversed three times has index 3 at every interior point

Example

Let γ(t)=exp(3it) for t[0,2π]. Then γ is a closed complex contour with trace the unit circle {z=1}, and

n(γ,z)=3  for z<1,n(γ,z)=0  for z>1.

The function λ(t)=3it is a continuous logarithm of γ0 along γ, its imaginary part θ(t)=3t is a continuous argument running from 0 to 6π, and (θ(2π)θ(0))/(2π)=3 recovers the index at the origin.

Facts & Assumptions

Given: The contour γ(t)=exp(3it) on [0,2π].

[L1]

For aC, r>0 and kZ, the contour γk(t)=a+rexp(ikt) on [0,2π] is a closed complex contour with n(γk,z)=k for za<r and n(γk,z)=0 for za>r; for k0 its trace is {z:za=r} (A circle traversed k times has winding number k inside and 0 outside).

[L2]

For a closed complex contour γ, a point p off its trace and a continuous argument θ of γp along γ, n(γ,p)=(θ(b)θ(a))/(2π) (The winding number is the increment of a continuous argument divided by 2π).

[L3]

n(γ,p)=(2πi)1γdz/(zp) (The winding number of a closed contour about a point off its trace).

[L4]

A continuous logarithm of γp along γ is a continuous λ with exp(λ(t))=γ(t)p for every t, and its continuous argument is Imλ (Continuous logarithms and continuous arguments along a contour).

Verification

technique · direct
1.1

Apply [L1] with a=0, r=1 and k=3: the contour is γ, it is a closed complex contour with trace {z=1}, and n(γ,z)=3 for z<1 while n(γ,z)=0 for z>1.

givenL1L3
1.2

The map λ(t)=3it is continuous on [0,2π] and satisfies exp(λ(t))=exp(3it)=γ(t)0, so it is a continuous logarithm of γ0 along γ in the sense of [L4], with continuous argument θ(t)=Im(3it)=3t by [L5].

givenL4L5
2.1

The argument increment is θ(2π)θ(0)=6π0=6π, so [L2] gives n(γ,0)=6π/(2π)=3, the same value step 1.1 assigns at the interior point 0.

step 1.1step 1.2L2

Depends on

Used by

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Sources