factorize-inside-C-x-p-x-1-i-x-n-n-1-i-x-n-n-

Question Number 30593 by abdo imad last updated on 23/Feb/18

f a c t o r i z e i n s i d e C [x] p (x) = {(1 + i \frac{x}{n})}^{n} - {(1 - i \frac{x}{n})}^{n} .

Answered by sma3l2996 last updated on 23/Feb/18

p (x) = {(1 + i x / n)}^{n} - {(1 - i x / n)}^{n}

= n^{- n} ({(n + i x)}^{n} - {(n - i x)}^{n})

= \frac{1}{n} \sqrt{n^{2} + x^{2}} ({(e^{i θ})}^{n} - {(e^{- i θ})}^{n}) ∖ θ = {t a n}^{- 1} (\frac{x}{n})

= \frac{\sqrt{n^{2} + x^{2}}}{n} (e^{i n θ} - e^{- i n θ}) = \frac{\sqrt{n^{2} + x^{2}}}{n} (2 i s i n (n θ))

p (x) = 2 i \frac{\sqrt{n^{2} + x^{2}}}{n} s i n ({n t a n}^{- 1} (\frac{x}{n}))

{t a n}^{- 1} (x_{k} / n) = \frac{k π}{n} \Rightarrow x_{k} = n t a n (\frac{k π}{n}) ∖ k = (0, 1, 2, \dots, n - 1)

s o r o o t s o f p (x) a r e α_{k} = n t a n (\frac{k π}{n})

s o p (x) = 2 i \sqrt{1 + {(\frac{x}{n})}^{2}} \prod_{k = 0}^{n - 1} (x - n t a n (\frac{k π}{n}))