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


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 39022 by maxmathsup by imad last updated on 01/Jul/18

$l e t p (x) = {(1 + e^{i θ} x)}^{n} - {(1 - e^{i θ} x)}^{n} w i t h n i n t e g r n a t u r a l$ $1) f i n d t h e r o o t s o f p (x)$ $2) f c t o r i z e i n s i d e C [x] p (x)$ $3) f a c t o r i z e i n s i d e R [x] p (x) . θ \in R$
Commented by math khazana by abdo last updated on 10/Jul/18
$1) let z =e^(iθ) x so?p(x)=0 ⇔ (((1−z)^n )/((1+z)^n )) =1⇔ (((1−z)/(1+z)))^n =1 ⇒((1−z_k )/(1+z_k )) = e^(i((kπ)/n)) k ∈[[0,n−1]] ⇒ 1−z_k =e^((ikπ)/n) + e^((ikπ)/n) z_k ⇒(1+e^((ikπ)/n) )z_k = 1−e^((ikπ)/n) ⇒ z_k = ((1−e^((ikπ)/n) )/(1+e^((ikπ)/n) )) = ((1−cos(((kπ)/n)) −i sin(((kπ)/n)))/(1+cos(((kπ)/n))+i sin(((kπ)/n)))) = ((2sin^2 (((kπ)/(2n)))−2isin(((kπ)/(2n)))cos(((kπ)/(2n))))/(2cos^2 (((kπ)/(2n))) +2i sin(((kπ)/(2n)))cos(((kπ)/(2n))))) =((−isin(((kπ)/(2n))) { cos(((kπ)/(2n))) +isin(((kπ)/(2n)))})/(cos(((kπ)/(2n))){ cos(((kπ)/(2n))) +isin(((kπ)/(2n)))})) =−i tan(((kπ)/(2n)))⇒ the roots of p(x) are the complex x_k =−i e^(−iθ) tan(((kπ)/(2n))) 2) p(x)=λ Π_(k=0) ^(n−1) (x−x_k ) =λ Π_(k=0) ^(n−1) (x +i e^(−iθ) tan(((kπ)/(2n)))) let determine λ wehave p(x)=(1+e^(iθ) x)^n −(1−e^(iθ) x)^n =Σ_(k=0) ^n C_n ^k e^(ikθ) x^k −Σ_(k=0) ^n C_n ^k (−1)^k e^(ikθ) x^k = Σ_(k=0) ^n C_n ^k (1−(−1)^k ) e^(ikθ) x^k = Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) 2 e^(i(2p+1)θ) x^(2p+1) ⇒ λ =2 C_n ^(2[((n−1)/2)]) e^(i{2[((n−1)/2)] +1)θ)$
$1) l e t z = e^{i θ} x s o ? p (x) = 0 \Leftrightarrow \frac{{(1 - z)}^{n}}{{(1 + z)}^{n}} = 1 \Leftrightarrow$ ${(\frac{1 - z}{1 + z})}^{n} = 1 \Rightarrow \frac{1 - z_{k}}{1 + z_{k}} = e^{i \frac{k π}{n}} k \in [[0, n - 1]] \Rightarrow$ $1 - z_{k} = e^{\frac{i k π}{n}} + e^{\frac{i k π}{n}} z_{k} \Rightarrow (1 + e^{\frac{i k π}{n}}) z_{k} = 1 - e^{\frac{i k π}{n}} \Rightarrow$ $z_{k} = \frac{1 - e^{\frac{i k π}{n}}}{1 + e^{\frac{i k π}{n}}} = \frac{1 - c o s (\frac{k π}{n}) - i s i n (\frac{k π}{n})}{1 + c o s (\frac{k π}{n}) + i s i n (\frac{k π}{n})}$ $= \frac{2 {s i n}^{2} (\frac{k π}{2 n}) - 2 i s i n (\frac{k π}{2 n}) c o s (\frac{k π}{2 n})}{2 {c o s}^{2} (\frac{k π}{2 n}) + 2 i s i n (\frac{k π}{2 n}) c o s (\frac{k π}{2 n})}$ $= \frac{- i s i n (\frac{k π}{2 n}) {c o s (\frac{k π}{2 n}) + i s i n (\frac{k π}{2 n})}}{c o s (\frac{k π}{2 n}) {c o s (\frac{k π}{2 n}) + i s i n (\frac{k π}{2 n})}}$ $= - i t a n (\frac{k π}{2 n}) \Rightarrow t h e r o o t s o f p (x) a r e t h e c o m p l e x$ $x_{k} = - i e^{- i θ} t a n (\frac{k π}{2 n})$ $2) p (x) = λ \prod_{k = 0}^{n - 1} (x - x_{k})$ $= λ \prod_{k = 0}^{n - 1} (x + i e^{- i θ} t a n (\frac{k π}{2 n})) l e t d e t e r m i n e λ$ $w e h a v e p (x) = {(1 + e^{i θ} x)}^{n} - {(1 - e^{i θ} x)}^{n}$ $= \sum_{k = 0}^{n} C_{n}^{k} e^{i k θ} x^{k} - \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} e^{i k θ} x^{k}$ $= \sum_{k = 0}^{n} C_{n}^{k} (1 - {(- 1)}^{k}) e^{i k θ} x^{k}$ $= \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} 2 e^{i (2 p + 1) θ} x^{2 p + 1} \Rightarrow$ $λ = 2 C_{n}^{2 [\frac{n - 1}{2}]} e^{i {2 [\frac{n - 1}{2}] + 1) θ}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum