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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-

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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) ?
$$\mathrm{If}\:\mathrm{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{satisfying}\: \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\:=\:\mathrm{0}\:,\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{x}^{\mathrm{53}} +\mathrm{x}^{\mathrm{52}} +\mathrm{x}^{\mathrm{51}} +\mathrm{x}^{\mathrm{50}} +\mathrm{x}^{\mathrm{49}} \:? \\ $$
Answered by john santu last updated on 09/Oct/20
we know that x^2 +x+1 = ((x^3 −1)/(x−1)) , x≠1 since x^2 +x+1= 0, implies that x^3 =1 and x≠1 . Now 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
$${we}\:{know}\:{that}\:{x}^{\mathrm{2}} +{x}+\mathrm{1}\:=\:\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}−\mathrm{1}}\:,\:{x}\neq\mathrm{1} \\ $$$${since}\:{x}^{\mathrm{2}} +{x}+\mathrm{1}=\:\mathrm{0},\:{implies}\:{that}\:{x}^{\mathrm{3}} =\mathrm{1} \\ $$$${and}\:{x}\neq\mathrm{1}\:.\:{Now}\:{x}^{\mathrm{53}} +{x}^{\mathrm{52}} +{x}^{\mathrm{51}} +{x}^{\mathrm{50}} +{x}^{\mathrm{49}} = \\ $$$${x}^{\mathrm{51}} \left({x}^{\mathrm{2}} +{x}\right)+{x}^{\mathrm{49}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:= \\ $$$${x}^{\mathrm{51}} \left(−\mathrm{1}\right)+{x}^{\mathrm{49}} \left(\mathrm{0}\right)\:=\:−{x}^{\mathrm{51}} =−\left({x}^{\mathrm{3}} \right)^{\mathrm{17}} \\ $$$$=\:−\left(\mathrm{1}\right)^{\mathrm{17}} \:=\:−\mathrm{1} \\ $$
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=((−1±i(√3))/2)=cos120°±isin120° Hence,by Mauvra′s formula we have x^(48) =cos(48.120°)±isin(48.120°) =cos(16.360°)±isin(16.360°) =1±0=1⇒P=−x^(48) =−1
$$\mathrm{P}=\mathrm{x}^{\mathrm{53}} +\mathrm{x}^{\mathrm{52}} +\mathrm{x}^{\mathrm{51}} +\mathrm{x}^{\mathrm{50}} +\mathrm{x}^{\mathrm{49}} \\ $$$$=\mathrm{x}^{\mathrm{53}} +\mathrm{x}^{\mathrm{52}} +\mathrm{x}^{\mathrm{51}} +\mathrm{x}^{\mathrm{50}} +\mathrm{x}^{\mathrm{49}} +\mathrm{x}^{\mathrm{48}} −\mathrm{x}^{\mathrm{48}} \\ $$$$=\mathrm{x}^{\mathrm{51}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)+\mathrm{x}^{\mathrm{48}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)−\mathrm{x}^{\mathrm{48}} \\ $$$$=−\mathrm{x}^{\mathrm{48}} \left(\mathrm{since}\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{0}\left(\mathrm{by}\:\mathrm{hypothesis}\right)\right) \\ $$$$\mathrm{From}\:\mathrm{the}\:\mathrm{hypothesis}\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{infer}\:\mathrm{x}=\frac{−\mathrm{1}\pm\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{cos120}°\pm\mathrm{isin120}° \\ $$$$\mathrm{Hence},\mathrm{by}\:\mathrm{Mauvra}'\mathrm{s}\:\mathrm{formula}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{x}^{\mathrm{48}} =\mathrm{cos}\left(\mathrm{48}.\mathrm{120}°\right)\pm\mathrm{isin}\left(\mathrm{48}.\mathrm{120}°\right) \\ $$$$=\mathrm{cos}\left(\mathrm{16}.\mathrm{360}°\right)\pm\mathrm{isin}\left(\mathrm{16}.\mathrm{360}°\right) \\ $$$$=\mathrm{1}\pm\mathrm{0}=\mathrm{1}\Rightarrow\mathrm{P}=−\mathrm{x}^{\mathrm{48}} =−\mathrm{1} \\ $$
Answered by floor(10²Eta[1]) last updated on 09/Oct/20
x^2 +x+1=0⇒x^2 =−x−1 x^3 =−x^2 −x=x+1−x=1 ⇒x^(3k) =1^k =1 ∀ k∈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 ⇒x^(53) +x^(52) +x^(51) +x^(50) +x^(49) =2(x^2 +x)+1=−2+1=−1
$$\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{x}^{\mathrm{2}} =−\mathrm{x}−\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{3}} =−\mathrm{x}^{\mathrm{2}} −\mathrm{x}=\mathrm{x}+\mathrm{1}−\mathrm{x}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{3k}} =\mathrm{1}^{\mathrm{k}} =\mathrm{1}\:\forall\:\mathrm{k}\in\mathbb{N} \\ $$$$\mathrm{x}^{\mathrm{53}} =\mathrm{x}^{\mathrm{51}} .\mathrm{x}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{52}} =\mathrm{x}^{\mathrm{51}} .\mathrm{x}=\mathrm{x} \\ $$$$\mathrm{x}^{\mathrm{51}} =\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{50}} =\mathrm{x}^{\mathrm{48}} .\mathrm{x}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{49}} =\mathrm{x}^{\mathrm{48}} .\mathrm{x}=\mathrm{x} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{53}} +\mathrm{x}^{\mathrm{52}} +\mathrm{x}^{\mathrm{51}} +\mathrm{x}^{\mathrm{50}} +\mathrm{x}^{\mathrm{49}} \\ $$$$=\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}\right)+\mathrm{1}=−\mathrm{2}+\mathrm{1}=−\mathrm{1} \\ $$
Answered by Olaf last updated on 09/Oct/20
x^2 +x+1 = 0 = ((1−x^3 )/(1−x)) ⇒ x^3 = 1 x^2 +x+1 = 0 ⇒ x+(1/x) = −1 x^(53) +x^(52) +x^(51) +x^(50) +x^(49) = x^(51) (x^2 +x+1+(1/x)+(1/x^2 )) = x^(51) [(x+(1/x))^2 −2+(x+(1/x))+1] = x^(51) (1−2−1+1) = −x^(51) = −(x^3 )^(17) = −1
$${x}^{\mathrm{2}} +{x}+\mathrm{1}\:=\:\mathrm{0}\:=\:\frac{\mathrm{1}−{x}^{\mathrm{3}} }{\mathrm{1}−{x}}\:\Rightarrow\:{x}^{\mathrm{3}} \:=\:\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{x}+\mathrm{1}\:=\:\mathrm{0}\:\Rightarrow\:{x}+\frac{\mathrm{1}}{{x}}\:=\:−\mathrm{1} \\ $$$${x}^{\mathrm{53}} +{x}^{\mathrm{52}} +{x}^{\mathrm{51}} +{x}^{\mathrm{50}} +{x}^{\mathrm{49}} \:= \\ $$$${x}^{\mathrm{51}} \left({x}^{\mathrm{2}} +{x}+\mathrm{1}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:= \\ $$$${x}^{\mathrm{51}} \left[\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}+\left({x}+\frac{\mathrm{1}}{{x}}\right)+\mathrm{1}\right]\:= \\ $$$${x}^{\mathrm{51}} \left(\mathrm{1}−\mathrm{2}−\mathrm{1}+\mathrm{1}\right)\:=\:−{x}^{\mathrm{51}} \:=\:−\left({x}^{\mathrm{3}} \right)^{\mathrm{17}} \:=\:−\mathrm{1} \\ $$

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