# Laurent Series, Singularities, and Residues *Local expansions around isolated singularities, the classification of those singularities, the residue theorem, and mechanical recipes for computing residues at simple and higher-order poles.* ## Laurent series A power series (Chapter 3) represents a function in a disk on which it is holomorphic. Around an isolated singularity we need the **Laurent series**, which admits negative powers. > **Theorem.** Let f be holomorphic on an annulus > `A = {z : r < |z − z₀| < R}`. Then f has a unique representation > > ``` > f(z) = Σ_{n = −∞}^{+∞} a_n (z − z₀)^n > ``` > > converging uniformly on compact subsets of A, with > > ``` > a_n = (1 / 2πi) ∮_γ f(z) / (z − z₀)^{n+1} dz > ``` > > for any positively oriented simple closed contour γ in A enclosing z₀. The `n ≥ 0` part is the **regular** part; the `n < 0` part is the **principal** part. The coefficient `a_{−1}` is distinguished enough to earn its own name: the **residue** of f at z₀, ``` Res(f; z₀) = a_{−1} = (1 / 2πi) ∮_γ f(z) dz ``` so the residue is already defined as "the thing you get from a closed contour integral around the singularity, divided by 2πi". ## Classification of isolated singularities Let z₀ be an isolated singularity of f — holomorphic on a punctured disk `0 < |z − z₀| < R` but not (as given) at z₀ itself. The principal part of the Laurent expansion classifies z₀: | Principal part | Name | Behaviour as z → z₀ | |---|---|---| | All `a_n = 0` for n < 0 | **Removable** | f bounded; define f(z₀) = a_0 to extend holomorphically | | Finitely many non-zero `a_n` for n < 0, largest |n| = m | **Pole of order m** | |f(z)| → ∞ as z → z₀; `(z − z₀)^m f(z)` is holomorphic and non-zero at z₀ | | Infinitely many non-zero `a_n` for n < 0 | **Essential singularity** | Values wild: Casorati–Weierstrass, Picard's great theorem | A pole of order 1 is called a **simple pole**. The canonical Laurent expansion of a simple pole is ``` f(z) = c_{−1}/(z − z₀) + a_0 + a_1 (z − z₀) + ... ``` with residue `c_{−1}`. **Example.** `f(z) = 1/(z² + 1) = 1/[(z − i)(z + i)]` has simple poles at z = ±i. Near z = i, ``` f(z) = 1 / [(z − i) · (z − i + 2i)] = (1/(z − i)) · 1/(2i + (z − i)) = (1/(2i)) · 1/(z − i) · 1/(1 + (z − i)/(2i)) ``` so `Res(f; i) = 1/(2i) = −i/2`. **Example.** `e^{1/z}` at z = 0: the Laurent series is `Σ_{n = 0}^{∞} z^{−n}/n!`, with infinitely many negative powers — an **essential** singularity. Residue `a_{−1} = 1`. ## The residue theorem Cauchy's theorem says closed integrals of holomorphic functions vanish. The residue theorem says **exactly how much** they fail to vanish when poles are enclosed. > **Residue theorem.** Let f be holomorphic on an open set D except at > isolated singularities, and let γ be a positively oriented simple > closed contour in D bounding a region inside which f has > singularities at z_1, ..., z_k (and no others). Then > > ``` > ∮_γ f(z) dz = 2πi Σ_{j = 1}^{k} Res(f; z_j) > ``` Proof sketch: surround each z_j with a small disjoint circle C_j in the positive orientation. The region bounded by γ externally and the C_j internally contains no singularities, so by Cauchy's theorem (with a cut to make it simply connected, or via a homotopy argument) the integral over the full boundary is zero: ``` ∮_γ f dz − Σ_j ∮_{C_j} f dz = 0 ``` Each `∮_{C_j} f dz` equals `2πi · Res(f; z_j)` by the definition of residue, giving the theorem. ## Computing residues — practical recipes **Simple pole, explicit factorisation.** If `f(z) = g(z)/(z − z₀)` with g holomorphic and non-zero at z₀, ``` Res(f; z₀) = g(z₀) ``` **Simple pole, quotient form.** If `f(z) = g(z)/h(z)` with g, h holomorphic, `g(z₀) ≠ 0`, `h(z₀) = 0`, and `h′(z₀) ≠ 0`, then ``` Res(f; z₀) = g(z₀) / h′(z₀) ``` Useful when h is a transcendental function whose zeros you know. **Example.** `Res(cot z; nπ)` for integer n: write `cot z = cos z / sin z`. Take g = cos, h = sin, so `h′ = cos`. At z = nπ, `g/h′ = cos(nπ)/cos(nπ) = 1`. Every integer multiple of π is a simple pole of cot with residue 1. **Pole of order m.** ``` Res(f; z₀) = (1 / (m − 1)!) lim_{z → z₀} d^{m−1}/dz^{m−1} [ (z − z₀)^m f(z) ] ``` **Example.** `f(z) = e^z / (z − 1)²` has a pole of order 2 at z = 1. ``` (z − 1)² f(z) = e^z d/dz [e^z] = e^z, at z = 1: e Res(f; 1) = (1/1!) · e = e ``` **Residue at infinity.** For f analytic outside some disc, the residue at infinity is ``` Res(f; ∞) = −(1 / 2πi) ∮_γ f(z) dz ``` where γ is a large positively oriented circle. Equivalently, with the substitution `w = 1/z`, `Res(f; ∞) = − Res(w^{−2} f(1/w); 0)`. Useful as a cross-check: the sum of **all** residues (finite and infinite) of a rational function is zero. ## The worked example that matters for K–K Evaluate the residue of ``` f(ω′) = χ(ω′) / (ω′ − ω) ``` at ω′ = ω, where χ is analytic at ω (real ω on the real axis, χ holomorphic in a neighbourhood). This is a simple pole with g(ω′) = χ(ω′), so ``` Res( χ(ω′)/(ω′ − ω) ; ω ) = χ(ω) ``` When we **indent** the K–K contour around ω′ = ω by a small upper semicircle of radius ε, the contour traverses **half** the full circle, so picks up **half** the residue (with a sign from the clockwise orientation of the indent relative to the region we exclude). Precisely: the limit ε → 0 of the integral over the indenting semicircle in the UHP (traversed clockwise, since we're cutting out a bite from the interior region of the big closed contour) is ``` lim_{ε → 0} ∫_{C_ε^{clockwise upper half}} χ(ω′)/(ω′ − ω) dω′ = −iπ χ(ω) ``` This "half residue" identity is the content of the Sokhotski–Plemelj formula (Chapter 8). ## Worked example: a real integral by residues A staple integral that illustrates the whole machinery. ``` I = ∫_0^{∞} dx / (1 + x²) ``` Extend to `(1/2) ∫_{−∞}^{∞} dx/(1 + x²)` (integrand even). Consider `f(z) = 1/(1 + z²) = 1/[(z − i)(z + i)]` with pole at z = i in the UHP. Close the contour with a large semicircle in the UHP. On the semicircle of radius R, |f(z)| ≤ 1/(R² − 1), and length πR, so by ML the semicircle contribution ≤ πR/(R² − 1) → 0 as R → ∞. The closed contour encloses one pole at z = i with residue `1/(2i) = −i/2`. Residue theorem: ``` ∫_{−∞}^{∞} dx/(1 + x²) = 2πi · (−i/2) = π ``` So `I = π/2`, matching the antiderivative `arctan x` evaluated at ∞. Every K–K calculation is a more elaborate version of exactly this pattern. ## See also - [05_contour_integration.md](05_contour_integration.md) - [06_cauchy_theorems.md](06_cauchy_theorems.md) - [08_principal_value.md](08_principal_value.md) - [09_jordan_lemma.md](09_jordan_lemma.md)