If-x-is-a-complex-number-satisfying-x-2-x-1-0-what-is-the-value-of-x-53-x-52-x-51-x-50-x-49-

Question Number 117045 by bobhans last updated on 09/Oct/20

If x is a complex number satisfying

x^{2} + x + 1 = 0, what is the value of

x^{53} + x^{52} + x^{51} + x^{50} + x^{49} ?

Answered by john santu last updated on 09/Oct/20

w e k n o w t h a t x^{2} + x + 1 = \frac{x^{3} - 1}{x - 1}, x \neq 1

s i n c e x^{2} + x + 1 = 0, i m p l i e s t h a t x^{3} = 1

a n d x \neq 1 . N o w x^{53} + x^{52} + x^{51} + x^{50} + x^{49} =

x^{51} (x^{2} + x) + x^{49} (x^{2} + x + 1) =

x^{51} (- 1) + x^{49} (0) = - x^{51} = - {(x^{3})}^{17}

= - {(1)}^{17} = - 1

Answered by 1549442205PVT last updated on 09/Oct/20

P = x^{53} + x^{52} + x^{51} + x^{50} + x^{49}

= x^{53} + x^{52} + x^{51} + x^{50} + x^{49} + x^{48} - x^{48}

= x^{51} (x^{2} + x + 1) + x^{48} (x^{2} + x + 1) - x^{48}

= - x^{48} (since x^{2} + x + 1 = 0 (by hypothesis))

From the hypothesis x^{2} + x + 1 = 0

we infer x = \frac{- 1 \pm i \sqrt{3}}{2} = \cos 120 ° \pm isin 120 °

Hence, by {Mauvra}^{'} s formula we have

x^{48} = \cos (48.120 °) \pm isin (48.120 °)

= \cos (16.360 °) \pm isin (16.360 °)

= 1 \pm 0 = 1 \Rightarrow P = - x^{48} = - 1

Answered by floor(10²Eta[1]) last updated on 09/Oct/20

x^{2} + x + 1 = 0 \Rightarrow x^{2} = - x - 1

x^{3} = - x^{2} - x = x + 1 - x = 1

\Rightarrow x^{3 k} = 1^{k} = 1 \forall k \in N

x^{53} = x^{51} . x^{2} = x^{2}

x^{52} = x^{51} . x = x

x^{51} = 1

x^{50} = x^{48} . x^{2} = x^{2}

x^{49} = x^{48} . x = x

\Rightarrow x^{53} + x^{52} + x^{51} + x^{50} + x^{49}

= 2 (x^{2} + x) + 1 = - 2 + 1 = - 1

Answered by Olaf last updated on 09/Oct/20

x^{2} + x + 1 = 0 = \frac{1 - x^{3}}{1 - x} \Rightarrow x^{3} = 1

x^{2} + x + 1 = 0 \Rightarrow x + \frac{1}{x} = - 1

x^{53} + x^{52} + x^{51} + x^{50} + x^{49} =

x^{51} (x^{2} + x + 1 + \frac{1}{x} + \frac{1}{x^{2}}) =

x^{51} [{(x + \frac{1}{x})}^{2} - 2 + (x + \frac{1}{x}) + 1] =

x^{51} (1 - 2 - 1 + 1) = - x^{51} = - {(x^{3})}^{17} = - 1

Facebook Tweet Pin