If $\alpha, \beta \in \mathbf{C}$ are the distinct roots, of the equation $x^{2}-x+1=0$, then $\alpha^{101}+\beta^{107}$ is equal to :

JEE Main · 2018 · 08 Apr 2018 · Offline · Complex Numbers

α, β \in C

are the distinct roots, of the equation

x^{2} - x + 1 = 0

, then

α^{101} + β^{107}

is equal to :

Answer: (D)

The roots of

x^{2} - x + 1 = 0

are

x = \frac{1 \pm i 3}{2}

, the primitive sixth roots of unity, equal to

- ω

and

- ω^{2}

where

ω = e^{2 π i /3}

satisfies

ω^{3} = 1

and

1 + ω + ω^{2} = 0

. Taking

α = - ω

and

β = - ω^{2}

α^{101} = (- ω)^{101} = - ω^{101} = - ω^{2}

because

101 \equiv 2 (mod 3)

, and

β^{107} = (- ω^{2})^{107} = - ω^{214} = - ω

because

214 \equiv 1 (mod 3)

. Their sum is

- (ω^{2} + ω) = - (- 1) = 1

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option