Contour ∫dθ/(√2-cosθ)

Method. Put z = e^{iθ}, so θ: 0 → 2π maps to the unit circle |z| = 1, with

dθ = dz/(iz), cos θ = (z² + 1)/(2z)

Transform:

√2 − cos θ = √2 − (z² + 1)/(2z) = (2√2 z − z² − 1)/(2z)

I = ∮_C [ 2z/(2√2 z − z² − 1) ] · dz/(iz) = ∮_C (2/i) dz/(−(z² − 2√2 z + 1)) = 2i ∮_C dz/(z² − 2√2 z + 1)

(since −2/i = 2i)

Poles: z² − 2√2 z + 1 = 0 gives z = √2 ± 1.

• z₁ = √2 − 1 ≈ 0.414 : inside C
• z₂ = √2 + 1 ≈ 2.414 : outside C

Residue at z₁:

Res = 1/(z₁ − z₂) = 1/((√2 − 1) − (√2 + 1)) = −1/2

Result: I = 2i · 2πi · (−1/2) = 2π.

Answer: ∫₀^{2π} dθ/(√2 − cos θ) = 2π

(Check: 2π/√(a² − b²) = 2π/√(2 − 1) = 2π.)