let-p-x-1-e-i-x-n-1-e-i-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-R-

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) 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}