let-p-x-1-jx-n-1-jx-n-1-find-the-roots-of-p-x-2-factorize-p-x-inside-C-x-j-e-i-2pi-3-

Question Number 35056 by math khazana by abdo last updated on 14/May/18

l e t p (x) = {(1 + j x)}^{n} - {(1 - j x)}^{n}

1) f i n d t h e r o o t s o f p (x)

2) f a c t o r i z e p (x) i n s i d e C [x]

j = e^{i \frac{2 π}{3}} .

Commented by math khazana by abdo last updated on 10/Jun/18

p (x) = 0 \Leftrightarrow \frac{{(1 - j x)}^{n}}{{(1 + j x)}^{n}} = 1 \Leftrightarrow {(\frac{1 - j x}{1 + j x})}^{n} = 1 t h e r o o t s

o f z^{n} = 1 a r e t h e c o m p l e x z_{k} = e^{i \frac{2 k π}{n}} k \in [[0, n - 1]]

s o 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 Z_{k} /

\frac{1 - {j Z}_{k}}{1 + {j Z}_{k}} = z_{k} \Leftrightarrow 1 - {j Z}_{k} = z_{k} + {j z}_{k} Z_{k} \Leftrightarrow

j (1 + z_{k}) Z_{k} = 1 - z_{k} \Leftrightarrow Z_{k} = \frac{1}{j} \frac{1 - z_{k}}{1 + z_{k}} \Rightarrow

j Z_{k} = \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{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 s i n (\frac{k π}{n}) e^{i \frac{k π}{n}}}{c o s (\frac{k π}{n}) e^{i \frac{k π}{n}}} = - i t a n (\frac{k π}{n}) \Rightarrow

Z_{k} = \frac{- i}{j} t a n (\frac{k π}{n}) = e^{- i \frac{π}{2}} e^{- i \frac{2 π}{3}} t a n (\frac{k π}{n})

= e^{- i (\frac{7 π}{6})} t a n (\frac{k π}{n}) w i t h n \in [[1, n - 1]]

Commented by math khazana by abdo last updated on 10/Jun/18

p (x) = λ \prod_{k = 1}^{n - 1} (x - Z_{k}) = λ \prod_{k = 1}^{n - 1} (x - e^{- i \frac{7 π}{6}} t a n (\frac{k π}{n}))

λ i s t h e d o m i n e n t c o e f f i c i e n t l e t f i n d i t

w e h a v e p (x) = {(1 + j x)}^{n} - {(1 - j x)}^{n}

= \sum_{k = 0}^{n} C_{n}^{k} j^{k} x^{k} - \sum_{k = 0}^{n} C_{n}^{k} {(- j)}^{k} x^{k}

= \sum_{k = 0}^{n} C_{n}^{k} (j^{k} - {(- j)}^{k}) x^{k} \Rightarrow

λ = C_{n}^{n} (j^{n} - {(- j)}^{n}) = e^{i \frac{2 n π}{3}} - {(- 1)}^{n} e^{i \frac{2 n π}{3}}

= (1 - {(- 1)}^{n}) e^{i \frac{2 n π}{3}} \Rightarrow

p (x) = {1 - {(- 1)}^{n}} e^{i \frac{2 n π}{3}} \prod_{k = 1}^{n} (x - e^{- i \frac{7 π}{6}} t a n (\frac{k π}{n})) .