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

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Complex Numbers

Let

α = \frac{- 1 + i 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:

Answer: (B)

Here

α = \frac{- 1 + i 3}{2} = ω

, a primitive cube root of unity with

ω^{3} = 1

and

1 + ω + ω^{2} = 0

. Then

b = \sum_{k = 0}^{100} (ω^{3})^{k} = \sum_{k = 0}^{100} 1 = 101

. For

a

, the geometric sum

\sum_{k = 0}^{100} ω^{2 k} = \frac{1 - ω ^{202}}{1 - ω ^{2}}

, and since

ω^{202} = ω

this is

\frac{1 - ω}{1 - ω ^{2}} = \frac{1}{1 + ω}

, so

a = (1 + ω) \cdot \frac{1}{1 + ω} = 1

. A quadratic with roots

1

and

101

x^{2} - (1 + 101) x + 101 = x^{2} - 102 x + 101 = 0

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option