Let \textrm{ }\textrm{ } \alpha = \frac{- 1 + i \sqrt{3}}{2} \textrm{ }\textrm{ } if \textrm{ }\textrm{ } a = \left(\right. 1 + \alpha \left.\right) \sum_{k = 0}^{100} \left(\alpha\right)^{2 k} \textrm{ }\textrm{ } and \textrm{ }\textrm{ } b = \sum_{k = 0}^{100} \left(\alpha\right)^{3 k}, then a and b are the roots of the quadratic equation:

Engineering

Mathematics

Cube Roots of Unity

Question

$Let α = \frac{- 1 + i \sqrt{3}}{2} if a = (1 + α) \sum_{k = 0}^{100} α^{2 k} and b = \sum_{k = 0}^{100} α^{3 k}$ , then a and b are the roots of the quadratic equation:

x² + 102x + 101 = 0

x² + 101x + 100 = 0

x² – 102x + 101 = 0

x² –101x + 100 = 0

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α = ω, b = 1 + ω³ + ω⁶ + ……= 101

a = (1 + ω) (1 + ω² + ω⁴ + …..ω¹⁹⁸ + ω²⁰⁰)

$= (1 + ω) \frac{(1 - {(ω^{2})}^{101})}{1 - ω^{2}} = \frac{(1 + ω) (1 - ω)}{1 - ω^{2}} = 1$

Equation : x² – (101 +1)x + (101) × 1 = 0 ⇒ x² – 102x + 101 = 0