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Question Number 94479 by i jagooll last updated on 19/May/20

find solution in C x^4 +x^3 +x^2 +x+1 = 0

$$\mathrm{find}\:\mathrm{solution}\:\mathrm{in}\:\mathbb{C}\: \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$

Answered by john santu last updated on 19/May/20

⇒(x−1)(x^4 +x^3 +x^2 +x+1) = 0 ⇒x^5 −1 = 0 x^5 = 1 ⇒ (x^5 )^(1/5) = (e^(2πi) )^(1/5) ⇒ x = e^((((2π+2πn)/5)).i) x_1 = cos (((2π)/5)) + i sin (((2π)/5)) x_2 = cos (((4π)/5)) + i sin (((4π)/5)) x_3 = cos (((6π)/5)) + i sin (((6π)/5)) x_4 = cos (((8π)/5)) + i sin (((8π)/5))

$$\Rightarrow\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{5}} \:−\mathrm{1}\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{5}} \:=\:\mathrm{1}\:\Rightarrow\:\left({x}^{\mathrm{5}} \right)^{\frac{\mathrm{1}}{\mathrm{5}}} \:=\:\left({e}^{\mathrm{2}\pi{i}} \right)^{\frac{\mathrm{1}}{\mathrm{5}}} \\ $$$$\Rightarrow\:{x}\:=\:{e}^{\left(\frac{\mathrm{2}\pi+\mathrm{2}\pi\mathrm{n}}{\mathrm{5}}\right).{i}} \\ $$$${x}_{\mathrm{1}} \:=\:\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:+\:{i}\:\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right) \\ $$$${x}_{\mathrm{2}} \:=\:\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:+\:{i}\:\mathrm{sin}\:\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right) \\ $$$${x}_{\mathrm{3}} \:=\:\mathrm{cos}\:\left(\frac{\mathrm{6}\pi}{\mathrm{5}}\right)\:+\:{i}\:\mathrm{sin}\:\left(\frac{\mathrm{6}\pi}{\mathrm{5}}\right) \\ $$$${x}_{\mathrm{4}} =\:\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{5}}\right)\:+\:{i}\:\mathrm{sin}\:\left(\frac{\mathrm{8}\pi}{\mathrm{5}}\right)\: \\ $$

Commented by i jagooll last updated on 19/May/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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