Taylor's Theorem, Laurent's Series & Complete Classification of Singularities and Poles
Idea in plain words
Taylor's Theorem: If f(z) is analytic within a disk |z - z₀| < R, it can be expanded into a unique power series: f(z) = ∑_{n=0}^∞ a_n (z - z₀)^n where a_n = f^{(n)}(z₀) / n!.
Exam tip
High-yield exam topic (Very High).
Explanation
Where Taylor's series is confined to disks of holomorphy free of singularities, Laurent's theorem handles functions punctured by singular points. The presence and length of the principal part uniquely diagnoses the nature of the singularity, separating tame removable points from poles and infinitely chaotic essential singularities.
Key points
- Laurent's Theorem: If f(z) is analytic in an annular region r < |z - z₀| < R, it has a unique expansion: f(z) = ∑{n=0}^∞ a_n (z - z₀)^n + ∑{n=1}^∞ b_n / (z - z₀)^n.
- Analytic Part: ∑_{n=0}^∞ a_n (z - z₀)^n (regular non-negative powers).
- Principal Part: ∑_{n=1}^∞ b_n / (z - z₀)^n (negative powers of z - z₀).
- Classification of Singularities based on Principal Part:
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- Removable Singularity: Principal part has ZERO terms (b_n = 0 for all n). lim_{z → z₀} f(z) exists and is finite. (e.g., (sin z) / z at z = 0).
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- Pole of Order m: Principal part terminates after m terms: b_m / (z - z₀)^m with b_m ≠ 0. If m = 1, it is a Simple Pole. lim_{z → z₀} f(z) = ∞.
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- Essential Singularity: Principal part has an INFINITE number of non-zero negative terms. lim_{z → z₀} f(z) does not exist (Picard's theorem). (e.g., e^{1/z} at z = 0).
Example
Example
Classify singularity of f(z) = (e^z - 1) / z: e^z - 1 = z + z²/2! + z³/3! + ... => f(z) = 1 + z/2! + z²/3! + ... No negative powers! z = 0 is a Removable Singularity.
Code examples
// Expansion in Different Annuli of f(z) = 1 / [ (z + 1)(z + 3) ]
// Partial Fractions:
// f(z) = [1/2] / (z + 1) - [1/2] / (z + 3)
//
// Region 1: Inside disk |z| < 1 (Taylor series):
// f(z) = (1/2) (1 + z)^-1 - (1/6) (1 + z/3)^-1
// = (1/2) ∑_{n=0}^∞ (-1)^n z^n - (1/6) ∑_{n=0}^∞ (-1)^n (z/3)^n
// = ∑_{n=0}^∞ (-1)^n [ 1/2 - 1/(2 * 3^{n+1}) ] z^n (Only positive powers)
//
// Region 2: Annulus 1 < |z| < 3 (Laurent series):
// For |z| > 1: (1 + z)^-1 = z^-1 (1 + 1/z)^-1 = ∑_{n=0}^∞ (-1)^n z^{-(n+1)}
// For |z| < 3: (z + 3)^-1 = (1/3)(1 + z/3)^-1 = ∑_{n=0}^∞ (-1)^n z^n / 3^{n+1}
// f(z) = (1/2) ∑_{n=0}^∞ (-1)^n / z^{n+1} - (1/2) ∑_{n=0}^∞ (-1)^n z^n / 3^{n+1}
// (Contains both positive and negative powers!)
//
// Region 3: Outside disk |z| > 3:
// f(z) = (1/2) [ 1/z (1 + 1/z)^-1 - 1/z (1 + 3/z)^-1 ]
// = (1/2) ∑_{n=0}^∞ (-1)^n [ 1 - 3^n ] / z^{n+1} (Only negative powers)
Exam questions
- [8 Marks] Expand f(z) = 1 / [(z - 1)(z - 2)] in Laurent series valid for: (i) |z| < 1, (ii) 1 < |z| < 2, (iii) |z| > 2, (iv) 0 < |z - 1| < 1.
- [7 Marks] Define isolated singularity, removable singularity, pole of order m, and essential singularity with an example of each.
- [7 Marks] Find and classify the singularities of f(z) = [z - sin z] / z³ and g(z) = e^{1/(z - 2)}.