Laurent Series Practice Worksheet | Complex Analysis

📚 Laurent Series Theory:

The Laurent series generalizes the Taylor series to include negative powers, essential for functions with singularities:

f(z) = ∑_n=-∞^∞ aₙ (z - z₀)ⁿ

Principal Part: Negative powers (n < 0) represent behavior near singularities.
Analytical Part: Non-negative powers (n ≥ 0) represent the analytic portion.

Problem 1: Expansion Around a Simple Pole Beginner

f(z) = e^z / z³

Find the Laurent series expansion centered at z = 0. Identify the singularity type and compute the residue.

Step 1: Recognize the expansion point

Singularity at z = 0 (pole of order 3)

Step 2: Expand e^z using Taylor series

e^z = 1 + z + z²/2! + z³/3! + z⁴/4! + z⁵/5! + ...

Step 3: Multiply by 1/z³

f(z) = (1/z³) + (1/z²) + (1/(2z)) + 1/6 + z/24 + z²/120 + ...

Step 4: Identify components

Principal Part: 1/z³ + 1/z² + 1/(2z)
Analytic Part: 1/6 + z/24 + z²/120 + ...
Residue (coefficient of 1/z): 1/2

💡 Key Insight: The residue at a pole of order m is found from the coefficient of (z - z₀)⁻¹ in the Laurent expansion.

Problem 2: Essential Singularity Intermediate

f(z) = e^{1/z} for z ≠ 0

Find the Laurent expansion about z = 0. What type of singularity is this?

Step 1: Substitute w = 1/z

f(z) = e^{1/z} = e^w where w = 1/z

Step 2: Use Maclaurin series for e^w

e^w = 1 + w + w²/2! + w³/3! + w⁴/4! + ...

Step 3: Substitute back w = 1/z

f(z) = 1 + (1/z) + (1/(2! z²)) + (1/(3! z³)) + (1/(4! z⁴)) + ...

Step 4: Analyze the series

Laurent Expansion: f(z) = ∑_n=0^∞ 1/(n! zⁿ)
Singularity Type: Essential singularity (infinite negative powers)
Residue: Coefficient of 1/z = 1

💡 Key Insight: Essential singularities have Laurent series with infinitely many negative powers.

Problem 3: Annulus Expansion Advanced

f(z) = 1/[(z-1)(z-2)]

Find Laurent expansions valid in: (a) 0 < |z-1| < 1 (b) 1 < |z-1| < ∞

Step 1: Partial fraction decomposition

1/[(z-1)(z-2)] = 1/(z-2) - 1/(z-1)

Step 2: Region (a): 0 < |z-1| < 1

For 1/(z-2): Write as -1/[1 - (z-1)] = -∑(z-1)ⁿ, n=0 to ∞ (converges for |z-1| < 1)

For -1/(z-1): Already in Laurent form

Combined: f(z) = -1/(z-1) - ∑(z-1)ⁿ, n=0 to ∞

Step 3: Region (b): 1 < |z-1| < ∞

For 1/(z-2): Write as 1/(z-1) × 1/[1 - 1/(z-1)] = 1/(z-1) ∑ 1/(z-1)ⁿ, n=0 to ∞

For -1/(z-1): Already in Laurent form

Combined: f(z) = ∑ 1/(z-1)ⁿ⁺², n=0 to ∞

💡 Key Insight: Laurent expansions depend on the annulus region - different regions yield different series!

📊 Laurent Series Quick Reference

Singularity Type	Laurent Series Pattern	Residue Extraction
Removable	No negative powers	Residue = 0
Pole of order m	Finite negative powers up to z⁻ᵐ	Coefficient of (z-z₀)⁻¹
Essential	Infinite negative powers	Coefficient of (z-z₀)⁻¹