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Question-179501

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Question Number 179501 by daus last updated on 30/Oct/22
Commented by Rasheed.Sindhi last updated on 30/Oct/22
Unclear!
$${Unclear}! \\ $$
Commented by daus last updated on 30/Oct/22
just help for z^4 −z^3 +2z^2 −z+1=0
$${just}\:{help}\:{for}\:\:{z}^{\mathrm{4}} −{z}^{\mathrm{3}} +\mathrm{2}{z}^{\mathrm{2}} −{z}+\mathrm{1}=\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 30/Oct/22
See the answer.
$${See}\:{the}\:{answer}. \\ $$
Answered by Rasheed.Sindhi last updated on 30/Oct/22
z=cosθ+isinθ z+(1/z)=cosθ+isinθ+(1/(cosθ+isinθ))∙((cosθ−isinθ)/(cosθ−isinθ)) =cosθ+isinθ+((cosθ−isinθ)/(cos^2 θ+sin^2 θ)) =cosθ+isinθ+cosθ−isinθ =2cosθ (z+(1/z))^2 =z^2 +(1/z^2 )+2=(2cosθ)^2 =4cos^2 θ z^2 −z+2−(1/z)+(1/z^2 ) =(z^2 +(1/z^2 ))−(z+(1/z))+2 =4cos^2 θ−2−2cosθ =4cos^2 θ−2cosθ−2 z^4 −z^3 +2z^2 −z+1=0 (z^4 /z^2 )−(z^3 /z^2 )+((2z^2 )/z^2 )−(z/z^2 )+(1/z^2 )=(0/z^2 ) z^2 −z+2−(1/z)+(1/z^2 )=0 (z^2 +(1/z^2 ))−(z+(1/z))+2=0 z+(1/z)=y⇒z^2 +(1/z^2 )=y^2 −2 y^2 −2−y+2=0 y(y−1)= y=0 ∣ y=1 z+(1/z)=0 ∣ z+(1/z)=1 z^2 +1= ∣ z^2 −z+1=0 z^2 =−1 ∣ z=((1±(√(1−4)))/2) z=±i ∣ z=((1±i(√3))/2)=(1/2)±((√3)/2)i Please verify the roots for validity.
$${z}=\mathrm{cos}\theta+{i}\mathrm{sin}\theta \\ $$$${z}+\frac{\mathrm{1}}{{z}}=\mathrm{cos}\theta+{i}\mathrm{sin}\theta+\frac{\mathrm{1}}{\mathrm{cos}\theta+{i}\mathrm{sin}\theta}\centerdot\frac{\mathrm{cos}\theta−{i}\mathrm{sin}\theta}{\mathrm{cos}\theta−{i}\mathrm{sin}\theta}\:\: \\ $$$$=\mathrm{cos}\theta+{i}\mathrm{sin}\theta+\frac{\mathrm{cos}\theta−{i}\mathrm{sin}\theta}{\mathrm{cos}^{\mathrm{2}} \theta+\mathrm{sin}^{\mathrm{2}} \theta} \\ $$$$=\mathrm{cos}\theta+\cancel{{i}\mathrm{sin}\theta}+\mathrm{cos}\theta−\cancel{{i}\mathrm{sin}\theta} \\ $$$$=\mathrm{2cos}\theta \\ $$$$\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} ={z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }+\mathrm{2}=\left(\mathrm{2cos}\theta\right)^{\mathrm{2}} =\mathrm{4cos}^{\mathrm{2}} \theta \\ $$$${z}^{\mathrm{2}} −{z}+\mathrm{2}−\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{z}^{\mathrm{2}} } \\ $$$$=\left({z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right)−\left({z}+\frac{\mathrm{1}}{{z}}\right)+\mathrm{2} \\ $$$$=\mathrm{4cos}^{\mathrm{2}} \theta−\mathrm{2}−\mathrm{2cos}\theta \\ $$$$=\mathrm{4cos}^{\mathrm{2}} \theta−\mathrm{2cos}\theta−\mathrm{2} \\ $$$${z}^{\mathrm{4}} −{z}^{\mathrm{3}} +\mathrm{2}{z}^{\mathrm{2}} −{z}+\mathrm{1}=\mathrm{0} \\ $$$$\frac{{z}^{\mathrm{4}} }{{z}^{\mathrm{2}} }−\frac{{z}^{\mathrm{3}} }{{z}^{\mathrm{2}} }+\frac{\mathrm{2}{z}^{\mathrm{2}} }{{z}^{\mathrm{2}} }−\frac{{z}}{{z}^{\mathrm{2}} }+\frac{\mathrm{1}}{{z}^{\mathrm{2}} }=\frac{\mathrm{0}}{{z}^{\mathrm{2}} } \\ $$$${z}^{\mathrm{2}} −{z}+\mathrm{2}−\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{z}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left({z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right)−\left({z}+\frac{\mathrm{1}}{{z}}\right)+\mathrm{2}=\mathrm{0} \\ $$$${z}+\frac{\mathrm{1}}{{z}}={y}\Rightarrow{z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }={y}^{\mathrm{2}} −\mathrm{2} \\ $$$${y}^{\mathrm{2}} −\mathrm{2}−{y}+\mathrm{2}=\mathrm{0} \\ $$$${y}\left({y}−\mathrm{1}\right)= \\ $$$${y}=\mathrm{0}\:\mid\:{y}=\mathrm{1} \\ $$$${z}+\frac{\mathrm{1}}{{z}}=\mathrm{0}\:\mid\:{z}+\frac{\mathrm{1}}{{z}}=\mathrm{1} \\ $$$${z}^{\mathrm{2}} +\mathrm{1}=\:\mid\:{z}^{\mathrm{2}} −{z}+\mathrm{1}=\mathrm{0} \\ $$$${z}^{\mathrm{2}} =−\mathrm{1}\:\mid\:{z}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}}}{\mathrm{2}} \\ $$$${z}=\pm{i}\:\mid\:{z}=\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i} \\ $$$${Please}\:{verify}\:{the}\:{roots}\:{for}\:{validity}. \\ $$

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