Taylor & Laurent Series, Singularities
Laurent series extend Taylor series with negative powers, describing functions near singularities. Learn complex Taylor series and convergence discs, Laurent series in annuli, principal part, classification of singularities (removable, poles of order m, essential), and residues as the 1/(z − z₀) coefficient.
Poles of transfer functions determine system dynamics; their residues give the weights of each mode in the time response. Laurent series are the precise tool for understanding behaviour around such points.
An analogy
A Laurent series is a Taylor series with a 'magnifying glass' on a singularity: the negative powers capture how violently the function blows up there.
Taylor in ℂ
Analytic functions equal their Taylor series in the largest disc avoiding singularities — the radius equals the distance to the nearest singularity (explains why 1/(1 + x²) has R = 1 on the real line).
Laurent series
Valid in annuli between singularities. Different annuli give different series for the same function. Build them with geometric-series tricks rather than coefficient integrals.
Singularities
Removable: no negative powers (sin z/z at 0). Pole of order m: finitely many (1/z²). Essential: infinitely many (e^(1/z)) — wild behaviour (Picard's theorem).
Why residues matter
Integrating a Laurent series term by term around z₀ kills every term except a₋₁(z − z₀)⁻¹, giving 2πj a₋₁ — the residue theorem.
Formulas
- Laurent series:
f(z) = Σ_{n=−∞}^{∞} aₙ(z − z₀)ⁿ - Principal part:
Σ_{n<0} aₙ(z − z₀)ⁿ - Pole of order m:
Principal part ends at a₋ₘ; (z − z₀)ᵐ f analytic, non-zero at z₀ - Geometric expansions:
1/(1 − w) = Σ wⁿ (|w| < 1)
Worked example: Laurent series in two annuli
Given: f = 1/(z(1 − z))
Find: Series for 0 < |z| < 1 and |z| > 1.
- 0 < |z| < 1: (1/z)Σzⁿ = 1/z + 1 + z + …
- |z| > 1: −1/z² · 1/(1 − 1/z) = −1/z² − 1/z³ − …
Answer: Residue at 0 is 1 (from the first series)
Worked example: Classify
Given: (1 − cos z)/z², e^(1/z), 1/(z − 2)³
Find: Singularity type.
- (1 − cos z)/z² = ½ − z²/24 + … → removable
- e^(1/z) = Σ z⁻ⁿ/n! → essential
- Pole of order 3 at 2
Answer: Removable, essential, pole of order 3
Common mistakes
- Wrong: Using the inner-annulus series outside its region. — Right: Each annulus has its own series.
- Wrong: Counting a removable singularity as a pole. — Right: Check the limit — if finite, it's removable.
Applications
- Transfer functions: Poles and partial-fraction residues.
- Z-transforms: Laurent series in z⁻¹ with regions of convergence.
- Asymptotics: Behaviour near singular points.
Key takeaways
- Laurent = Taylor + negative powers.
- Series depend on the annulus.
- Removable, pole, essential.
- Residue = a₋₁.
Going further
- Picard's theorem: Essential singularities hit almost every value.
- Branch points: Non-isolated singularities (√z, log z).
- Mittag-Leffler: Functions from their poles.