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Question Number 36904 by prof Abdo imad last updated on 07/Jun/18

$$1)decomposeinsideC\left[x\right]$$$$p\left(x\right)=x^{2n}-2\left(cos\alpha \right)x^n+1$$$$2)decoposep\left(x\right)insideR\left[x\right]$$

Commented by math khazana by abdo last updated on 11/Jun/18

$$1)letput{x}^n=t\Rightarrow p\left(x\right)=t^2-2cos\left(\alpha \right)t+1$$$${\mathrm{\Delta }}^{{}^{\prime }}={cos}^2-1={\left(isin\alpha \right)}^2\Rightarrow t_1=e^{i\alpha }and{t}_2=e^{-i\alpha }$$$$p\left(x\right)=(t-e^{i\alpha })(t-e^{-i\alpha })=(x^n-e^{i\alpha })(x^n-e^{-i\alpha })roots$$$$of{z}^n-e^{i\alpha }=0\Rightarrow z^n=e^{i\alpha }\Rightarrow r=1and$$$$n\theta =\alpha +2k\pi \Rightarrow \theta =\frac{\alpha +2k\pi }{n}k\in \left[[0,n-1]\right]so$$$$therootsare{z}_k=e^{i\frac{\alpha +2k\pi }{n}}$$$$rootsof{x}^n-e^{-i\alpha }=0\Rightarrow x^n=e^{-i\alpha }\Rightarrow r=1$$$$andn\theta =-\alpha +2k\pi \Rightarrow {\theta }_k=\frac{-\alpha +2k\pi }{n}\Rightarrow$$$$therootsare{\lambda }_k=e^{i\frac{-\alpha +2k\pi }{n}}andk\in \left[[0,n-1]\right]$$$$\Rightarrow p\left(x\right)=\prod _{k=0}^{n-1}(x-z_k)\prod _{k=0}^{n-1}(x-{\lambda }_k)$$$$=\prod _{k=0}^{n-1}(x-e^{i\frac{\alpha +2k\pi }{n}})(x-e^{i\frac{-\alpha +2k\pi }{n}})with$$$$k\in \left[[0,n-1]\right].$$

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