Question Number 39022 by maxmathsup by imad last updated on 01/Jul/18
![let p(x)= (1+e^(iθ) x)^n −(1−e^(iθ) x)^n with n integr natural 1) find the roots of p(x) 2) fctorize inside C[x] p(x) 3) factorize inside R[x] p(x). θ ∈R](Q39022.png)
$${let}\:{p}\left({x}\right)=\:\left(\mathrm{1}+{e}^{{i}\theta} {x}\right)^{{n}} \:−\left(\mathrm{1}−{e}^{{i}\theta} {x}\right)^{{n}} \:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{fctorize}\:{inside}\:{C}\left[{x}\right]\:{p}\left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{factorize}\:{inside}\:{R}\left[{x}\right]\:{p}\left({x}\right).\:\:\theta\:\in{R} \\ $$
Commented by math khazana by abdo last updated on 10/Jul/18
![1) let z =e^(iθ) x so?p(x)=0 ⇔ (((1−z)^n )/((1+z)^n )) =1⇔ (((1−z)/(1+z)))^n =1 ⇒((1−z_k )/(1+z_k )) = e^(i((kπ)/n)) k ∈[[0,n−1]] ⇒ 1−z_k =e^((ikπ)/n) + e^((ikπ)/n) z_k ⇒(1+e^((ikπ)/n) )z_k = 1−e^((ikπ)/n) ⇒ z_k = ((1−e^((ikπ)/n) )/(1+e^((ikπ)/n) )) = ((1−cos(((kπ)/n)) −i sin(((kπ)/n)))/(1+cos(((kπ)/n))+i sin(((kπ)/n)))) = ((2sin^2 (((kπ)/(2n)))−2isin(((kπ)/(2n)))cos(((kπ)/(2n))))/(2cos^2 (((kπ)/(2n))) +2i sin(((kπ)/(2n)))cos(((kπ)/(2n))))) =((−isin(((kπ)/(2n))) { cos(((kπ)/(2n))) +isin(((kπ)/(2n)))})/(cos(((kπ)/(2n))){ cos(((kπ)/(2n))) +isin(((kπ)/(2n)))})) =−i tan(((kπ)/(2n)))⇒ the roots of p(x) are the complex x_k =−i e^(−iθ) tan(((kπ)/(2n))) 2) p(x)=λ Π_(k=0) ^(n−1) (x−x_k ) =λ Π_(k=0) ^(n−1) (x +i e^(−iθ) tan(((kπ)/(2n)))) let determine λ wehave p(x)=(1+e^(iθ) x)^n −(1−e^(iθ) x)^n =Σ_(k=0) ^n C_n ^k e^(ikθ) x^k −Σ_(k=0) ^n C_n ^k (−1)^k e^(ikθ) x^k = Σ_(k=0) ^n C_n ^k (1−(−1)^k ) e^(ikθ) x^k = Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) 2 e^(i(2p+1)θ) x^(2p+1) ⇒ λ =2 C_n ^(2[((n−1)/2)]) e^(i{2[((n−1)/2)] +1)θ)](Q39740.png)
$$\left.\mathrm{1}\right)\:{let}\:\:{z}\:={e}^{{i}\theta} {x}\:\:{so}?{p}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\:\frac{\left(\mathrm{1}−{z}\right)^{{n}} }{\left(\mathrm{1}+{z}\right)^{{n}} }\:=\mathrm{1}\Leftrightarrow \\ $$$$\left(\frac{\mathrm{1}−{z}}{\mathrm{1}+{z}}\right)^{{n}} \:=\mathrm{1}\:\Rightarrow\frac{\mathrm{1}−{z}_{{k}} }{\mathrm{1}+{z}_{{k}} }\:=\:{e}^{{i}\frac{{k}\pi}{{n}}} \:\:\:\:\:{k}\:\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:\Rightarrow \\ $$$$\mathrm{1}−{z}_{{k}} ={e}^{\frac{{ik}\pi}{{n}}} \:+\:{e}^{\frac{{ik}\pi}{{n}}} \:{z}_{{k}} \:\:\Rightarrow\left(\mathrm{1}+{e}^{\frac{{ik}\pi}{{n}}} \right){z}_{{k}} =\:\mathrm{1}−{e}^{\frac{{ik}\pi}{{n}}} \:\Rightarrow \\ $$$${z}_{{k}} =\:\frac{\mathrm{1}−{e}^{\frac{{ik}\pi}{{n}}} }{\mathrm{1}+{e}^{\frac{{ik}\pi}{{n}}} }\:=\:\frac{\mathrm{1}−{cos}\left(\frac{{k}\pi}{{n}}\right)\:−{i}\:{sin}\left(\frac{{k}\pi}{{n}}\right)}{\mathrm{1}+{cos}\left(\frac{{k}\pi}{{n}}\right)+{i}\:{sin}\left(\frac{{k}\pi}{{n}}\right)} \\ $$$$=\:\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}}\right)−\mathrm{2}{isin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right){cos}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\:+\mathrm{2}{i}\:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right){cos}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)} \\ $$$$=\frac{−{isin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\:\left\{\:{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\:+{isin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\right\}}{{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\left\{\:{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\:+{isin}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\right\}} \\ $$$$=−{i}\:{tan}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\Rightarrow\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:{are}\:{the}\:{complex} \\ $$$${x}_{{k}} =−{i}\:{e}^{−{i}\theta} \:{tan}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right) \\ $$$$\left.\mathrm{2}\right)\:{p}\left({x}\right)=\lambda\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{x}_{{k}} \right) \\ $$$$=\lambda\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({x}\:+{i}\:{e}^{−{i}\theta} \:{tan}\left(\frac{{k}\pi}{\mathrm{2}{n}}\right)\right)\:{let}\:{determine}\:\lambda \\ $$$${wehave}\:{p}\left({x}\right)=\left(\mathrm{1}+{e}^{{i}\theta} {x}\right)^{{n}} \:−\left(\mathrm{1}−{e}^{{i}\theta} {x}\right)^{{n}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{e}^{{ik}\theta} \:{x}^{{k}} \:−\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\left(−\mathrm{1}\right)^{{k}} \:{e}^{{ik}\theta} \:{x}^{{k}} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{k}} \right)\:{e}^{{ik}\theta} \:{x}^{{k}} \\ $$$$=\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:\:\mathrm{2}\:{e}^{{i}\left(\mathrm{2}{p}+\mathrm{1}\right)\theta} \:{x}^{\mathrm{2}{p}+\mathrm{1}} \:\Rightarrow \\ $$$$\lambda\:=\mathrm{2}\:{C}_{{n}} ^{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\:{e}^{{i}\left\{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]\:+\mathrm{1}\right)\theta} \:\: \\ $$$$ \\ $$