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Every contour missing a point admits a continuous logarithm, unique up to a constant in 2πiZ

Statement

Let γ:[a,b]C be a complex contour and let pC with pγ. Then:

  1. there is a continuous logarithm λ of γp along γ (Continuous logarithms and continuous arguments along a contour);
  2. if λ1 and λ2 are two of them, then λ1λ2 is a constant function with value in 2πiZ;
  3. for each vC with expv=γ(a)p there is exactly one continuous logarithm λ of γp along γ with λ(a)=v.

In particular the increment λ(b)λ(a), and the increment θ(b)θ(a) of the associated continuous argument, are the same for every choice of λ. No differentiability of γ is used.

Facts & Assumptions

Given: A complex contour γ:[a,b]C and a point pγ.

[L1]

A continuous logarithm of γp along γ is a continuous λ:[a,b]C with exp(λ(t))=γ(t)p for every t; a holomorphic logarithm branch of zp on an open V missing p is a holomorphic L on V with exp(L(z))=zp (Continuous logarithms and continuous arguments along a contour).

[L2]

For a complex contour γ and pγ, the distance d=inf{wp:wγ} is positive and there is δ>0 such that every partition a=t0<<tr=b of mesh below δ has γ([ti,ti+1])D(γ(ti),d) and pD(γ(ti),d) for every i<r; at least one such partition exists (A contour missing a point subdivides into arcs lying in discs that miss it).

[L3]

If D(c,ρ) is an open disc with ρ>0 and pD(c,ρ), there is a holomorphic L on D(c,ρ) with exp(L(z))=zp there (A disc missing p carries a holomorphic logarithm of zp).

[L4]

ker(exp)=2πiZ, and expz=expw exactly when zw2πiZ (ker(exp)=2πiZ, and expz=expw exactly when zw2πiZ).

[L5]

The complex exponential maps C onto C{0} (The complex exponential maps C onto C{0}).

[L6]

exp(z+w)=expzexpw for all z,wC (exp(z+w)=expzexpw, and the complex exponential extends the real exponential).

[L7]

The continuous image of a connected subset is a connected subset (A continuous image of a connected space is connected, and connectedness is a topological property).

[L9]

If a property holds at 0 and passes from n to n+1, it holds for every natural number (The principle of mathematical induction).

[L10]

A composite of continuous maps is continuous, and a function whose restrictions to the members of a finite closed cover are continuous is continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous).

[L11]

For z=a+bi with a,b real, Imz=b and z=a2+b2 (Real and imaginary parts, complex conjugation, and modulus).

[L12]

A function complex differentiable at a point is continuous there (Complex differentiability at a point implies continuity there).

[L13]

The integers form an ordered commutative ring, and their canonical image in R is discrete; hence if m<n then m+12 lies strictly between them and is not an integer (The integers form a commutative ring, The integers form a totally ordered ring, Integer part: for every real x there is exactly one integer m with mx<m+1).

Proof

technique · direct
1.1

Since pγ, the number γ(a)p is nonzero, so [L5] supplies vC with expv=γ(a)p; more generally, for every such v the set of complex numbers with that exponential is v+2πiZ by [L4].

givenL4L5
1.2

If λ1,λ2 are continuous logarithms of γp along γ, then exp(λ1(t))=exp(λ2(t)) for every t, so λ1(t)λ2(t)2πiZ by [L4]; the real-valued function g=Im(λ1λ2)/(2π) is continuous by [L10] and [L11] and takes values in Z, so by [L7] and [L8] its image is an order-convex subset of R inside Z, which by [L13] can only be a single point. Hence λ1λ2 is a constant in 2πiZ, and it is 0 when λ1(a)=λ2(a).

L1L4L7L8L10L11L13
1.3

Assume a<b. By [L2] there are d>0 and a partition a=t0<t1<<tr=b with γ([ti,ti+1])Di:=D(γ(ti),d) and pDi for every i<r.

givenL2
2.1

By [L3] each Di carries a holomorphic Li with exp(Li(z))=zp for zDi, and Li is continuous on Di by [L12].

step 1.3L3L12
2.2

Define c0:=v and, for each i<r, define ci+1:=ci+Li(γ(ti+1))Li(γ(ti)); this determines the finite list c0,,cr. Now define λ on [ti,ti+1] by λ(t)=ci+Li(γ(t))Li(γ(ti)). The two formulas available at a shared point ti with 0<i<r agree, the ith giving ci and the (i1)st giving ci1+Li1(γ(ti))Li1(γ(ti1))=ci, so λ:[a,b]C is a well-defined function with λ(a)=v and λ(ti)=ci for every ir.

step 1.1step 1.3
3.1

Each restriction λ[ti,ti+1] is continuous, being a constant plus the composite of γ with Li of step 2.1; the intervals [ti,ti+1] form a finite closed cover of [a,b], so λ is continuous by [L10].

step 2.1step 2.2L10
3.2

For every i<r and t[ti,ti+1], [L6] gives exp(λ(t))=exp(ci)exp(Li(γ(t)))exp(Li(γ(ti)))1, and exp(Li(γ(t)))=γ(t)p by step 2.1, so exp(ci)=γ(ti)p forces exp(λ(t))=γ(t)p and, at t=ti+1, exp(ci+1)=γ(ti+1)p. Since exp(c0)=exp(v)=γ(a)p, an induction on i ([L9]) gives exp(λ(t))=γ(t)p for every t[a,b].

step 1.1step 2.1step 2.2L6L9
4.1

Steps 3.1 and 3.2 make λ a continuous logarithm of γp along γ with λ(a)=v, which proves claims 1 and 3 when a<b; when a=b the constant function with value v does the same, since its only value satisfies expv=γ(a)p. Claim 2 is step 1.2, which also gives the uniqueness in claim 3, and it makes λ(b)λ(a) and its imaginary part independent of the choice by [L1] and [L11].

step 1.2step 3.1step 3.2L1L11

Depends on

Used by

Cited to discharge well-definedness by Continuous logarithms and continuous arguments along a contour.

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Sources