let j =e^((i2π)/3) and p(x) =(1+xj)^n −(1−xj)^n 1) find the roots of p(x) and factorize inside C[x] p(x) 2) decompose inside C(x) the fration F(x) =(1/(p(x))) .


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Question Number 42507 by maxmathsup by imad last updated on 26/Aug/18

$l e t j = e^{\frac{i 2 π}{3}} a n d p (x) = {(1 + x j)}^{n} - {(1 - x j)}^{n}$ $1) f i n d t h e r o o t s o f p (x) a n d f a c t o r i z e i n s i d e C [x] p (x)$ $2) d e c o m p o s e i n s i d e C (x) t h e f r a t i o n F (x) = \frac{1}{p (x)} .$
Commented by maxmathsup by imad last updated on 28/Aug/18
$1) p(x)=0 ⇔ (((1+xj)/(1−xj)))^n =1 ⇒ ((1+xj)/(1−xj)) =e^((i2kπ)/n) k∈[[0,n−1]] let z_k =e^((i2kπ)/n) ⇒ ((1+xj)/(1−xj)) =z_k ⇒1+jx = z_k −jz_k x ⇒j(1+z_k )x = z_k −1 ⇒x =((z_k −1)/(j(1+z_k ))) =−(1/j) ((1−z_k )/(1+z_k )) ⇒ the roots are x_k =−(1/j) ((1−cos(((2kπ)/n))−isin(((2kπ)/n)))/(1+cos(((2kπ)/n))+isin(((2kπ)/n)))) =−(1/j) ((2sin^2 (((kπ)/n))−2isin(((kπ)/n))cos(((kπ)/n)))/(2cos^2 (((kπ)/n))+2i sin(((kπ)/n))cos(((kπ)/n)))) =(i/j) ((sin(((kπ)/n))(cos(((kπ)/n)) +isin(((kπ)/n))))/(cos(((kπ)/n))(cos(((kπ)/n))+isin(((kπ)/n))))) =(i/j)tan(((kπ)/n)) ⇒ x_k =(i/j) tan(((kπ)/n)) and 0≤k≤n−1 p(x) =λ Π_(k=0) ^(n−1) (x−x_k ) =λ Π_(k=0) ^(n−1) (x−(i/j)tan(((kπ)/n))) let findλ we have p(x) =Σ_(k=0) ^n C_n ^k j^k x^k −Σ_(k=0) ^(n−1) C_n ^k (−1)^k j^k x^k =Σ_(k=0) ^n C_n ^k {1−(−1)^k }j^k x^k ⇒λ = (1−(−1)^n )j^n ⇒ p(x) =(1−(−1)^n )j^n Π_(k=0) ^(n−1) (x−(i/j)tan(((kπ)/n))) .$
$1) p (x) = 0 \Leftrightarrow {(\frac{1 + x j}{1 - x j})}^{n} = 1 \Rightarrow \frac{1 + x j}{1 - x j} = e^{\frac{i 2 k π}{n}} k \in [[0, n - 1]] l e t z_{k} = e^{\frac{i 2 k π}{n}} \Rightarrow$ $\frac{1 + x j}{1 - x j} = z_{k} \Rightarrow 1 + j x = z_{k} - {j z}_{k} x \Rightarrow j (1 + z_{k}) x = z_{k} - 1 \Rightarrow x = \frac{z_{k} - 1}{j (1 + z_{k})}$ $= - \frac{1}{j} \frac{1 - z_{k}}{1 + z_{k}} \Rightarrow t h e r o o t s a r e x_{k} = - \frac{1}{j} \frac{1 - c o s (\frac{2 k π}{n}) - i s i n (\frac{2 k π}{n})}{1 + c o s (\frac{2 k π}{n}) + i s i n (\frac{2 k π}{n})}$ $= - \frac{1}{j} \frac{2 {s i n}^{2} (\frac{k π}{n}) - 2 i s i n (\frac{k π}{n}) c o s (\frac{k π}{n})}{2 {c o s}^{2} (\frac{k π}{n}) + 2 i s i n (\frac{k π}{n}) c o s (\frac{k π}{n})} = \frac{i}{j} \frac{s i n (\frac{k π}{n}) (c o s (\frac{k π}{n}) + i s i n (\frac{k π}{n}))}{c o s (\frac{k π}{n}) (c o s (\frac{k π}{n}) + i s i n (\frac{k π}{n}))}$ $= \frac{i}{j} t a n (\frac{k π}{n}) \Rightarrow x_{k} = \frac{i}{j} t a n (\frac{k π}{n}) a n d 0 ⩽ k ⩽ n - 1$ $p (x) = λ \prod_{k = 0}^{n - 1} (x - x_{k}) = λ \prod_{k = 0}^{n - 1} (x - \frac{i}{j} t a n (\frac{k π}{n})) l e t f i n d λ$ $w e h a v e p (x) = \sum_{k = 0}^{n} C_{n}^{k} j^{k} x^{k} - \sum_{k = 0}^{n - 1} C_{n}^{k} {(- 1)}^{k} j^{k} x^{k}$ $= \sum_{k = 0}^{n} C_{n}^{k} {1 - {(- 1)}^{k}} j^{k} x^{k} \Rightarrow λ = (1 - {(- 1)}^{n}) j^{n} \Rightarrow$ $p (x) = (1 - {(- 1)}^{n}) j^{n} \prod_{k = 0}^{n - 1} (x - \frac{i}{j} t a n (\frac{k π}{n})) .$
Commented by maxmathsup by imad last updated on 28/Aug/18
$2) we have p(x) =λ Π_(k=0) ^(n−1) (x−x_k ) with x_k =(i/j) tan(((kπ)/n)) ⇒ F(x)=(1/(p(x))) = (1/(λ Π_(k=0) ^(n−1) (x−x_k ))) = Σ_(k=0) ^(n−1) (α_k /(x−x_k )) (the poles are simples) α_k =(1/(p^′ (x_k ))) but p(x) =(1+jx)^n −(1−jx)^n ⇒p^′ (x)=nj(1+jx)^(n−1) +nj(1−jx)^(n−1) ⇒p^′ (x_k ) =nj(1 +jx_k )^(n−1) +nj(1−jx_k )^(n−1) ⇒ α_k =(1/(nj{ (1+jx_k )^(n−1) +(1−jx_k )^(n−1) })) .$
$2) w e h a v e p (x) = λ \prod_{k = 0}^{n - 1} (x - x_{k}) w i t h x_{k} = \frac{i}{j} t a n (\frac{k π}{n}) \Rightarrow$ $F (x) = \frac{1}{p (x)} = \frac{1}{λ \prod_{k = 0}^{n - 1} (x - x_{k})} = \sum_{k = 0}^{n - 1} \frac{α_{k}}{x - x_{k}} (t h e p o l e s a r e s i m p l e s)$ $α_{k} = \frac{1}{p^{^{'}} (x_{k})} b u t p (x) = {(1 + j x)}^{n} - {(1 - j x)}^{n} \Rightarrow p^{^{'}} (x) = n j {(1 + j x)}^{n - 1} + n j {(1 - j x)}^{n - 1}$ $\Rightarrow p^{^{'}} (x_{k}) = n j {(1 + {j x}_{k})}^{n - 1} + n j {(1 - {j x}_{k})}^{n - 1} \Rightarrow$ $α_{k} = \frac{1}{n j {{(1 + {j x}_{k})}^{n - 1} + {(1 - {j x}_{k})}^{n - 1}}} .$
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