Question Number 36904 by prof Abdo imad last updated on 07/Jun/18
![1)decompose inside C[x] p(x)=x^(2n) −2(cosα)x^n +1 2) decopose p(x)inside R[x]](Q36904.png)
$$\left.\mathrm{1}\right){decompose}\:{inside}\:{C}\left[{x}\right] \\ $$$${p}\left({x}\right)={x}^{\mathrm{2}{n}} \:−\mathrm{2}\left({cos}\alpha\right){x}^{{n}} \:+\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{decopose}\:{p}\left({x}\right){inside}\:{R}\left[{x}\right] \\ $$
Commented by math khazana by abdo last updated on 11/Jun/18
![1) let put x^n =t ⇒p(x)=t^2 −2cos(α)t +1 Δ^′ =cos^2 −1 =(isinα)^2 ⇒t_1 = e^(iα) and t_2 =e^(−iα) p(x)=(t−e^(iα) )(t−e^(−iα) ) = (x^n −e^(iα) )(x^n −e^(−iα) )roots of z^n −e^(iα) =0 ⇒ z^n =e^(iα) ⇒ r=1 and nθ = α +2kπ ⇒ θ =((α +2kπ)/n) k∈[[0,n−1]]so the roots are z_k =e^(i ((α+2kπ)/n)) roots of x^n −e^(−iα) =0 ⇒x^n =e^(−iα) ⇒r=1 and nθ =−α +2kπ ⇒ θ_k =((−α +2kπ)/n) ⇒ the roots are λ_k = e^(i((−α+2kπ)/n)) and k∈[[0,n−1]] ⇒p(x) =Π_(k=0) ^(n−1) (x−z_k )Π_(k=0) ^(n−1) (x−λ_k ) =Π_(k=0) ^(n−1) (x −e^(i((α+2kπ)/n)) )(x− e^(i((−α +2kπ)/n)) ) with k∈[[0,n−1]].](Q37240.png)
$$\left.\mathrm{1}\right)\:{let}\:{put}\:\:{x}^{{n}} \:={t}\:\Rightarrow{p}\left({x}\right)={t}^{\mathrm{2}} \:−\mathrm{2}{cos}\left(\alpha\right){t}\:+\mathrm{1} \\ $$$$\Delta^{'} \:={cos}^{\mathrm{2}} −\mathrm{1}\:=\left({isin}\alpha\right)^{\mathrm{2}} \:\Rightarrow{t}_{\mathrm{1}} =\:{e}^{{i}\alpha} \:{and}\:{t}_{\mathrm{2}} ={e}^{−{i}\alpha} \\ $$$${p}\left({x}\right)=\left({t}−{e}^{{i}\alpha} \right)\left({t}−{e}^{−{i}\alpha} \right)\:=\:\left({x}^{{n}} \:−{e}^{{i}\alpha} \right)\left({x}^{{n}} −{e}^{−{i}\alpha} \right){roots} \\ $$$${of}\:{z}^{{n}} \:−{e}^{{i}\alpha} \:\:\:=\mathrm{0}\:\Rightarrow\:{z}^{{n}} \:={e}^{{i}\alpha} \:\Rightarrow\:{r}=\mathrm{1}\:{and} \\ $$$${n}\theta\:=\:\alpha\:+\mathrm{2}{k}\pi\:\Rightarrow\:\theta\:=\frac{\alpha\:+\mathrm{2}{k}\pi}{{n}}\:\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]{so} \\ $$$${the}\:{roots}\:{are}\:{z}_{{k}} ={e}^{{i}\:\frac{\alpha+\mathrm{2}{k}\pi}{{n}}} \\ $$$${roots}\:{of}\:\:\:{x}^{{n}} \:−{e}^{−{i}\alpha} =\mathrm{0}\:\Rightarrow{x}^{{n}} \:={e}^{−{i}\alpha} \:\Rightarrow{r}=\mathrm{1} \\ $$$${and}\:{n}\theta\:=−\alpha\:+\mathrm{2}{k}\pi\:\Rightarrow\:\theta_{{k}} =\frac{−\alpha\:+\mathrm{2}{k}\pi}{{n}}\:\Rightarrow \\ $$$${the}\:{roots}\:{are}\:\:\lambda_{{k}} \:=\:{e}^{{i}\frac{−\alpha+\mathrm{2}{k}\pi}{{n}}} \:\:{and}\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$$\Rightarrow{p}\left({x}\right)\:=\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({x}−{z}_{{k}} \right)\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−\lambda_{{k}} \right) \\ $$$$=\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}\:−{e}^{{i}\frac{\alpha+\mathrm{2}{k}\pi}{{n}}} \right)\left({x}−\:{e}^{{i}\frac{−\alpha\:+\mathrm{2}{k}\pi}{{n}}} \right)\:{with} \\ $$$${k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]. \\ $$