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Every contour missing a point admits a continuous logarithm, unique up to a constant in
Statement
Let be a complex contour and let with . Then:
- there is a continuous logarithm of along (Continuous logarithms and continuous arguments along a contour);
- if and are two of them, then is a constant function with value in ;
- for each with there is exactly one continuous logarithm of along with .
In particular the increment , and the increment of the associated continuous argument, are the same for every choice of . No differentiability of is used.
Facts & Assumptions
Given: A complex contour and a point .
A continuous logarithm of along is a continuous with for every ; a holomorphic logarithm branch of on an open missing is a holomorphic on with (Continuous logarithms and continuous arguments along a contour).
For a complex contour and , the distance is positive and there is such that every partition of mesh below has and for every ; at least one such partition exists (A contour missing a point subdivides into arcs lying in discs that miss it).
If is an open disc with and , there is a holomorphic on with there (A disc missing carries a holomorphic logarithm of ).
, and exactly when (, and exactly when ).
The complex exponential maps onto (The complex exponential maps onto ).
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).
A subset is connected exactly when it is order-convex (The connected subspaces of with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in "); a closed bounded interval is order-convex (Intervals of : the nine order-convex forms, nondegeneracy, and length).
If a property holds at and passes from to , it holds for every natural number (The principle of mathematical induction).
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).
For with real, and (Real and imaginary parts, complex conjugation, and modulus).
A function complex differentiable at a point is continuous there (Complex differentiability at a point implies continuity there).
The integers form an ordered commutative ring, and their canonical image in is discrete; hence if then 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 there is exactly one integer with ).
Proof
Since , the number is nonzero, so [L5] supplies with ; more generally, for every such the set of complex numbers with that exponential is by [L4].
If are continuous logarithms of along , then for every , so by [L4]; the real-valued function is continuous by [L10] and [L11] and takes values in , so by [L7] and [L8] its image is an order-convex subset of inside , which by [L13] can only be a single point. Hence is a constant in , and it is when .
Assume . By [L2] there are and a partition with and for every .
By [L3] each carries a holomorphic with for , and is continuous on by [L12].
Define and, for each , define ; this determines the finite list . Now define on by . The two formulas available at a shared point with agree, the th giving and the st giving , so is a well-defined function with and for every .
Each restriction is continuous, being a constant plus the composite of with of step 2.1; the intervals form a finite closed cover of , so is continuous by [L10].
For every and , [L6] gives , and by step 2.1, so forces and, at , . Since , an induction on ([L9]) gives for every .
Steps 3.1 and 3.2 make a continuous logarithm of along with , which proves claims 1 and 3 when ; when the constant function with value does the same, since its only value satisfies . Claim 2 is step 1.2, which also gives the uniqueness in claim 3, and it makes and its imaginary part independent of the choice by [L1] and [L11].
Depends on
- Continuous logarithms and continuous arguments along a contour
- A contour missing a point subdivides into arcs lying in discs that miss it
- A disc missing $p$ carries a holomorphic logarithm of $z-p$
- $\ker(\exp)=2\pi i\mathbb Z$, and $\exp z=\exp w$ exactly when $z-w\in2\pi i\mathbb Z$
- The complex exponential maps $\mathbb C$ onto $\mathbb C\setminus\{0\}$
- $\exp(z+w)=\exp z\,\exp w$, and the complex exponential extends the real exponential
- A continuous image of a connected space is connected, and connectedness is a topological property
- The connected subspaces of $\mathbb{R}$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $\mathbb{R}$"
- The principle of mathematical induction
- Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous
- Real and imaginary parts, complex conjugation, and modulus
- Complex differentiability at a point implies continuity there
- The integers as equivalence classes of pairs of naturals
- 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 $m \le x < m + 1$
- Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length
Used by
- The winding number is the increment of a continuous argument divided by 2π Corollary
- A continuous argument computed along a spiralling contour Example
- The index of a cycle about a point off its trace is an integer Theorem
- The integral of dz/(z-p) along a contour is the increment of a continuous logarithm Theorem
- The winding number of a closed contour is an integer Theorem
Cited to discharge well-definedness by Continuous logarithms and continuous arguments along a contour.
Dependency tree · two levels
105 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
- L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 §2.1, Exercise 2 (standard reference, not scraped)
- J. Lebl, Complex Analysis, Ch. 4 §4.1 (standard reference, not scraped)