Question Number 92839 by i jagooll last updated on 09/May/20
$$\mathrm{let}\:\mathrm{a}\:\mathrm{is}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{such}\: \\ $$$$\mathrm{that}\:\mathrm{a}^{\mathrm{10}} \:+\:\mathrm{a}^{\mathrm{5}} \:+\mathrm{1}\:=\:\mathrm{0}. \\ $$$$\mathrm{find}\:\mathrm{a}^{\mathrm{2005}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2005}} }\:? \\ $$
Answered by john santu last updated on 09/May/20
![set a^5 = w ⇒ w^2 +w+1 = 0 w^2 +w+1 = ((w^3 −1)/(w−1)) = 0 ⇒a^(10) +a^5 +1 = ((a^(15) −1)/(a^5 −1)) = 0 a^(15) = 1 ⇒ since 2005 = 133×15+10 a^(2005) = (a^(15) )^(133) ×a^(10) = 1×a^(10) = a^(10) then a^(2005) +(1/a^(2005) ) = a^(10) +(1/a^(10) ) = a^(10) +((a^(15) )/a^(10) ) = a^(10) +a^5 = −1 [ a^(10) +a^5 +1 = 0 ]](https://www.tinkutara.com/question/Q92841.png)
$$\mathrm{set}\:{a}^{\mathrm{5}} \:=\:{w}\:\Rightarrow\:{w}^{\mathrm{2}} +{w}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${w}^{\mathrm{2}} +{w}+\mathrm{1}\:=\:\frac{{w}^{\mathrm{3}} −\mathrm{1}}{{w}−\mathrm{1}}\:=\:\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{10}} +{a}^{\mathrm{5}} +\mathrm{1}\:=\:\frac{{a}^{\mathrm{15}} −\mathrm{1}}{{a}^{\mathrm{5}} −\mathrm{1}}\:=\:\mathrm{0} \\ $$$${a}^{\mathrm{15}} \:=\:\mathrm{1}\:\Rightarrow\:\mathrm{since}\:\mathrm{2005}\:=\:\mathrm{133}×\mathrm{15}+\mathrm{10} \\ $$$${a}^{\mathrm{2005}} \:=\:\left({a}^{\mathrm{15}} \right)^{\mathrm{133}} \:×{a}^{\mathrm{10}} \:=\:\mathrm{1}×{a}^{\mathrm{10}} \:=\:{a}^{\mathrm{10}} \\ $$$$\mathrm{then}\:{a}^{\mathrm{2005}} \:+\frac{\mathrm{1}}{{a}^{\mathrm{2005}} }\:=\:{a}^{\mathrm{10}} +\frac{\mathrm{1}}{{a}^{\mathrm{10}} } \\ $$$$=\:{a}^{\mathrm{10}} \:+\frac{{a}^{\mathrm{15}} \:}{{a}^{\mathrm{10}} }\:=\:{a}^{\mathrm{10}} +{a}^{\mathrm{5}} \\ $$$$=\:−\mathrm{1}\:\:\left[\:{a}^{\mathrm{10}} +{a}^{\mathrm{5}} +\mathrm{1}\:\:=\:\mathrm{0}\:\right]\: \\ $$
Commented by i jagooll last updated on 09/May/20
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by peter frank last updated on 09/May/20
$${thank}\:{you} \\ $$