let give the polynomial P(x)= (1/(2i))( (1+ix)^n −(1−ix)^n ) .find the roots of P(x) and factorize P(x).


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Question Number 28267 by abdo imad last updated on 22/Jan/18

$l e t g i v e t h e p o l y n o m i a l$ $P (x) = \frac{1}{2 i} ({(1 + i x)}^{n} - {(1 - i x)}^{n}) . f i n d t h e r o o t s o f P (x)$ $a n d f a c t o r i z e P (x) .$
Commented by abdo imad last updated on 23/Jan/18

$P (x) = 0 {(1 + i x)}^{n} = {(1 - i x)}^{n} \Leftrightarrow {(\frac{1 - i x}{1 + i x})}^{n} = 1 l e t p u t$ $\frac{1 - i x}{1 + i x} = z (e) \Leftrightarrow z^{n} = 1 w i c h h a v e f o r r o o t s$ $z_{k} = e^{i \frac{2 k π}{n}} a n d k \in [[0, n - 1]] s o t h e r o o t s o f P (x) a r e t h e$ $c o m p l e x e s x_{k} / \frac{1 - x_{k}}{1 + x_{k}} = z_{k} \Leftrightarrow 1 - x_{k} = z_{k} + z_{k} x_{k}$ $\Leftrightarrow (1 + z_{k}) x_{k} = 1 - z_{k} \Leftrightarrow x_{k} = \frac{1 - z_{k}}{1 + z_{k}} = \frac{1 - c o s (\frac{2 k π}{n}) - i s i n (\frac{2 k π}{n})}{1 + c o s (\frac{2 k π}{n}) + i s i n (\frac{2 k π}{n})}$ $= \frac{2 {s i n}^{2} (\frac{k π}{n}) - 2 i s i n (\frac{k π}{n}) c o s (\frac{k π}{n})}{2 {c o s}^{2} (\frac{k π}{n}) + 2 i s i n (\frac{k π}{n}) c o s (\frac{k π}{n})}$ $= \frac{- i s i n (\frac{k π}{n}) e^{i \frac{π}{n}}}{c o s (\frac{k π}{n}) e^{i \frac{π}{n}}} = - i t a n (\frac{k π}{n}) s o t h e r o o t s o f P (x) a r e$ $x_{k} = - i t a n (\frac{k π}{n}) a n d k \in [[0, n - 1]]$ $P (x) = λ \prod_{k = 0}^{n - 1} (x + i t a n (\frac{k π}{n})) l e t f i n d λ ?$ $w e h a v e 2 i P (x) = \sum_{k = 0}^{n} C_{n}^{k} {(i x)}^{k} - \sum_{k = 0}^{n} C_{n}^{k} {(- i x)}^{k}$ $= \sum_{k = 0}^{n} C_{n}^{k} (i^{k} - {(- i)}^{k}) x^{k} = 2 i \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} {(- 1)}^{p} x^{2 p + 1}$ $\Rightarrow P (x) = \sum_{p = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{p} C_{n}^{2 p + 1} x^{2 p + 1} a n d$ $λ = {(- 1)}^{[\frac{n - 1}{2}]} C_{n}^{2 [\frac{n - 1}{2}] + 1} f i n a l l y$ $P (x) = {(- 1)}^{[\frac{n - 1}{2}]} C_{n}^{2 [\frac{n - 1}{2}] + 1} \prod_{k = 0}^{k = n - 1} (x + i t a n (\frac{k π}{n})) .$
	Answered by sma3l2996 last updated on 23/Jan/18

	$P (x) = \frac{1}{2 i} ({(1 + i x)}^{n} - {(1 - i x)}^{n})$ ${l e t}^{'} s p u t z = 1 + i x = \sqrt{1 + x^{2}} e^{i θ} w i t h θ = {t a n}^{- 1} x$ $s o P (x) = \frac{1}{2 i} ({(\sqrt{1 + x^{2}} e^{i θ})}^{n} - {(\sqrt{1 + x^{2}} e^{- i θ})}^{n})$ $= \frac{{(\sqrt{1 + x^{2}})}^{n}}{2 i} (e^{i n θ} - e^{- i n θ}) = {(\sqrt{1 + x^{2}})}^{n} s i n (n θ)$ $P (x) = {(1 + x^{2})}^{n / 2} s i n ({n t a n}^{- 1} (x))$
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