= cosθ ± i sinθ ∴ α = cosθ + i sinθ and β = cosθ - i sinθ ∴ αn = (cosθ+isinθ)n = cosnθ + i sinnθ and βn = (cosθ−isinθ)n = cosnθ - i sinnθ [using De-Moivre's theorem] Now, αn+βn = 2 cos nθ and αn.βn = 1 ∴ Equation having αn and βn as roots is x2−(αn+βn)x+αnβn = 0 ⇒ x2−2cosnθx+1=0