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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 \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) 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}}} .