Calculation of Residues, Cauchy's Residue Theorem & Evaluation of Real Trigonometric Integrals

Explanation

Cauchy's Residue Theorem is the pinnacle of complex analysis. By expressing definite real integrals as closed contours around the unit circle |z| = 1, difficult integration is reduced to algebraic polynomial root-finding and residue calculation at poles situated strictly inside the unit disk.

Key points

• Residue Formula for Simple Pole (order 1): Res(z₀) = lim_{z → z₀} (z - z₀) f(z).
• Alternative Simple Pole Formula: If f(z) = P(z) / Q(z) where P(z₀) ≠ 0 and Q(z₀) = 0, Q'(z₀) ≠ 0, then Res(z₀) = P(z₀) / Q'(z₀).
• Residue Formula for Pole of Order m: Res(z₀) = (1 / (m - 1)!) lim_{z → z₀} [ d^{m-1}/dz^{m-1} { (z - z₀)^m f(z) } ].
• Cauchy's Residue Theorem: ∮C f(z) dz = 2πi ∑{k=1}^N Res[f(z), z_k], where z_k are all isolated poles enclosed inside simple closed contour C.
• Evaluation of Real Trigonometric Integrals: I = ∫_0^{2π} R(cos θ, sin θ) dθ.
• Transformation to unit circle C: |z| = 1 via z = e^{iθ}:
• dz = i e^{iθ} dθ = i z dθ => dθ = dz / (iz).
• cos θ = (z + 1/z) / 2 = (z² + 1) / (2z); sin θ = (z - 1/z) / (2i) = (z² - 1) / (2iz).

Example

Code examples

// Complete Contour Evaluation: Real Trigonometric Integral
// Evaluate: I = ∫_0^{2π} dθ / (a + b cos θ)   (where a > b > 0)
// Step 1: Map to unit circle C: |z| = 1 using z = e^{iθ}:
//    cos θ = (z^2 + 1) / (2z),  dθ = dz / (i z)
//    I = ∮_C [ dz / (i z) ] / [ a + b(z^2 + 1)/(2z) ]
//      = ∮_C [ dz / (i z) ] / [ (2az + b z^2 + b) / (2z) ]
//      = (2 / i) ∮_C dz / [ b z^2 + 2az + b ]
//      = (2 / (i b)) ∮_C dz / [ z^2 + (2a/b) z + 1 ]
//
// Step 2: Find poles (roots of z^2 + (2a/b) z + 1 = 0):
//    z = [ -(2a/b) ± √(4a^2/b^2 - 4) ] / 2
//      = -(a/b) ± √((a^2 - b^2)/b^2) = [ -a ± √(a^2 - b^2) ] / b
//    Let α = [ -a + √(a^2 - b^2) ] / b  and  β = [ -a - √(a^2 - b^2) ] / b
//    Notice: α · β = 1.
//    Since a > b > 0: |β| > 1 (lies OUTSIDE |z| = 1)
//    Therefore: |α| < 1 (lies strictly INSIDE |z| = 1!)
//
// Step 3: Compute Residue at simple pole z = α:
//    f(z) = 1 / [ (z - α)(z - β) ]
//    Res(α) = lim_{z -> α} (z - α) f(z) = 1 / (α - β)
//    α - β = 2 √(a^2 - b^2) / b
//    Res(α) = b / [ 2 √(a^2 - b^2) ]
//
// Step 4: Apply Cauchy's Residue Theorem:
//    I = (2 / (i b)) * [ 2πi * Res(α) ]
//      = (2 / (i b)) * [ 2πi * (b / (2 √(a^2 - b^2))) ]
//      = 2π / √(a^2 - b^2)  (Exact University Analytical Solution!)

Exam questions

• [8 Marks] State Cauchy's Residue Theorem. Using contour integration, evaluate ∫_0^{2π} [dθ / (5 - 3 cos θ)].
• [8 Marks] Evaluate ∫_0^{2π} [cos 2θ / (5 + 4 cos θ)] dθ using residue calculus.
• [7 Marks] Find the residues of f(z) = z² / [(z - 1)² (z + 2)] at all its poles.