Question Number 69794 by mathmax by abdo last updated on 27/Sep/19
![let p(x)=(x+1)^6 −e^(iα) with α real 1) find the roots of p(x) 2) factorize p(x)inside C[x] 3)factorize p(x)inside R[x]](https://www.tinkutara.com/question/Q69794.png)
$${let}\:{p}\left({x}\right)=\left({x}+\mathrm{1}\right)^{\mathrm{6}} \:−{e}^{{i}\alpha} \:\:\:\:{with}\:\alpha\:{real} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{p}\left({x}\right){inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{3}\right){factorize}\:{p}\left({x}\right){inside}\:{R}\left[{x}\right] \\ $$
Commented by mathmax by abdo last updated on 01/Oct/19
![1)p(x)=0 ⇔(x+1)^6 =e^(iα) ⇔(((x+1)^6 )/e^(iα) ) =1 ⇔(((x+1)^6 )/((e^((iα)/6) )^6 )) =1 ⇒ ((x+1)e^(−((iα)/6)) )^6 =e^(i2kπ) ⇒(x+1)e^(−((iα)/6)) =e^((i2kπ)/6) ⇒ (x+1)=e^((ikπ)/3) e^((iα)/6) ⇒x+1 =e^((i(2kπ +α))/6) ⇒x=e^(i(((2kπ+α))/6)) −1 so the roots of p(x) are z_k =e^(i(((2kπ+α)/6))) −1 with k∈[[0,5]] 2) p(x) =aΠ_(k=0) ^5 (x−z_k ) we see that a=1 ⇒ p(x) =Π_(k=0) ^5 (x+1−e^(i(((3kπ+α)/6))) )](https://www.tinkutara.com/question/Q70141.png)
$$\left.\mathrm{1}\right){p}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\left({x}+\mathrm{1}\right)^{\mathrm{6}} ={e}^{{i}\alpha} \:\Leftrightarrow\frac{\left({x}+\mathrm{1}\right)^{\mathrm{6}} }{{e}^{{i}\alpha} }\:=\mathrm{1}\:\Leftrightarrow\frac{\left({x}+\mathrm{1}\right)^{\mathrm{6}} }{\left({e}^{\frac{{i}\alpha}{\mathrm{6}}} \right)^{\mathrm{6}} }\:=\mathrm{1}\:\Rightarrow \\ $$$$\left(\left({x}+\mathrm{1}\right){e}^{−\frac{{i}\alpha}{\mathrm{6}}} \right)^{\mathrm{6}} \:={e}^{{i}\mathrm{2}{k}\pi} \:\Rightarrow\left({x}+\mathrm{1}\right){e}^{−\frac{{i}\alpha}{\mathrm{6}}} \:={e}^{\frac{{i}\mathrm{2}{k}\pi}{\mathrm{6}}} \:\Rightarrow \\ $$$$\left({x}+\mathrm{1}\right)={e}^{\frac{{ik}\pi}{\mathrm{3}}} \:{e}^{\frac{{i}\alpha}{\mathrm{6}}} \:\Rightarrow{x}+\mathrm{1}\:={e}^{\frac{{i}\left(\mathrm{2}{k}\pi\:+\alpha\right)}{\mathrm{6}}} \:\Rightarrow{x}={e}^{{i}\left(\frac{\left.\mathrm{2}{k}\pi+\alpha\right)}{\mathrm{6}}\right.} −\mathrm{1} \\ $$$${so}\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:{are}\:{z}_{{k}} ={e}^{{i}\left(\frac{\mathrm{2}{k}\pi+\alpha}{\mathrm{6}}\right)} −\mathrm{1}\:\:{with}\:{k}\in\left[\left[\mathrm{0},\mathrm{5}\right]\right] \\ $$$$\left.\mathrm{2}\right)\:{p}\left({x}\right)\:={a}\prod_{{k}=\mathrm{0}} ^{\mathrm{5}} \left({x}−{z}_{{k}} \right)\:\:\:{we}\:{see}\:{that}\:{a}=\mathrm{1}\:\Rightarrow \\ $$$${p}\left({x}\right)\:=\prod_{{k}=\mathrm{0}} ^{\mathrm{5}} \left({x}+\mathrm{1}−{e}^{{i}\left(\frac{\mathrm{3}{k}\pi+\alpha}{\mathrm{6}}\right)} \right) \\ $$
Commented by mathmax by abdo last updated on 01/Oct/19
$${error}\:{of}\:{typo}\:\:\:\:{p}\left({x}\right)=\prod_{{k}=\mathrm{0}} ^{\mathrm{5}} \left({x}+\mathrm{1}−{e}^{{i}\left(\frac{\mathrm{2}{k}\pi+\alpha}{\mathrm{6}}\right)} \right) \\ $$
Answered by mind is power last updated on 29/Sep/19
![p(x)=0⇒ (x+1)^6 =e^(ia) ⇔(x+1)^6 =e^(i(a+2kπ)) ⇒x+1=e^(i(a+2kπ)/6) ,k∈[0,5] ⇒x_k =−1+e^(i((a+2kπ)/6)) 2 )p(x)=Π_(k=0) ^5 (x−(−1+e^(i((a+2kπ)/6)) )) p(x)=Π_(k=0) ^5 (x+1−e^(i((a+2kπ)/6)) ) over IR not possible only[if e^(ia) ∈IR⇒a=kπ cause p(−1)=−e^(ia)](https://www.tinkutara.com/question/Q69981.png)
$${p}\left({x}\right)=\mathrm{0}\Rightarrow \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{6}} ={e}^{{ia}} \Leftrightarrow\left({x}+\mathrm{1}\right)^{\mathrm{6}} ={e}^{{i}\left({a}+\mathrm{2}{k}\pi\right)} \\ $$$$\Rightarrow{x}+\mathrm{1}={e}^{{i}\left({a}+\mathrm{2}{k}\pi\right)/\mathrm{6}} ,{k}\in\left[\mathrm{0},\mathrm{5}\right] \\ $$$$\Rightarrow{x}_{{k}} =−\mathrm{1}+{e}^{{i}\frac{{a}+\mathrm{2}{k}\pi}{\mathrm{6}}} \\ $$$$\left.\mathrm{2}\:\right){p}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\prod}}\left({x}−\left(−\mathrm{1}+{e}^{{i}\frac{{a}+\mathrm{2}{k}\pi}{\mathrm{6}}} \right)\right) \\ $$$${p}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\prod}}\left({x}+\mathrm{1}−{e}^{{i}\frac{{a}+\mathrm{2}{k}\pi}{\mathrm{6}}} \right) \\ $$$${over}\:{IR}\:{not}\:{possible}\:{only}\left[{if}\:{e}^{{ia}} \in{IR}\Rightarrow{a}={k}\pi\right. \\ $$$${cause}\:{p}\left(−\mathrm{1}\right)=−{e}^{{ia}} \\ $$$$ \\ $$$$ \\ $$