let p(x)= (1+ix)^n −(1−ix)^n 1) find the roots of p(x) and factorize p(x) ) give p(x) at form of arcs.


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 30522 by abdo imad last updated on 22/Feb/18

$l e t p (x) = {(1 + i x)}^{n} - {(1 - i x)}^{n}$ $1) f i n d t h e r o o t s o f p (x) a n d f a c t o r i z e p (x)$ $) g i v e p (x) a t f o r m o f a r c s .$
Commented by abdo imad last updated on 24/Feb/18

$1) w e h a v e p (x) = 2 i I m {(1 + i x)}^{n} b u t ∣ 1 + i x ∣= \sqrt{1 + x^{2}}$ $\Rightarrow 1 + i x = \sqrt{1 + x^{2}} (\frac{1}{\sqrt{1 + x^{2}}} + i \frac{x}{\sqrt{1 + x^{2}}}) = r e^{i θ} \Rightarrow r = \sqrt{1 + x^{2}}$ $a n d c o s θ = \frac{1}{\sqrt{1 + x^{2}}} a n d s i n θ = \frac{x}{\sqrt{1 + x^{2}}} \Rightarrow t a n θ = x$ $\Rightarrow θ = a r t a n x \Rightarrow {(1 + i x)}^{n} = (\sqrt{1 + x^{2})}^{n} e^{i n a r c t a n x} \Rightarrow$ $p (x) = 2 i {(\sqrt{1 + x^{2}})}^{n} s i n (n a r c t a n x)$ $p (x) = 0 \Leftrightarrow n a r c t a n x = k π \Leftrightarrow a r c t a n x = \frac{k π}{n} \Leftrightarrow x_{k} = t a n (\frac{k π}{n})$ $a n d k \in [[0, n - 1]] a n d p (x) = λ \prod_{k = 0}^{n - 1} (x - t a n (\frac{k π}{n})) l e t$ $f i n d λ w e h a v e p (x) = \sum_{k = 0}^{n} C_{n}^{k} {(i x)}^{k} - \sum_{k = 0}^{n} C_{n}^{k} {(- i x)}^{k}$ $= \sum_{k = 0}^{n} C_{n}^{k} ({(i)}^{k} - {(- i)}^{k}) x^{k} = 2 i \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} x^{2 p + 1}$ $\Rightarrow λ = 2 i C_{n}^{2 [\frac{n - 1}{2}] + 1} \Rightarrow$ $p (x) = 2 i C_{n}^{2 [\frac{n - 1}{2}] + 1} \prod_{k = 0}^{n - 1} (x - t a n (\frac{k π}{n})) .$ $2) p (x) = 2 i {(\sqrt{1 + x^{2}})}^{n} s i n (n a r c t a n x) .$
	Answered by sma3l2996 last updated on 23/Feb/18
	$p(x)=(√((1+x^2 )^n ))(e^(inθ) −e^(−inθ) ) \θ=tan^(−1) x =(√((1+x^2 )^n ))(2isin(nθ)) p(x)=2i(√((1+x^2 )^n ))sin(ntan^(−1) (x)) tan^(−1) (x_k )=((kπ)/n) ⇒x_k =tan(((kπ)/n)) so roots of p(x) are α_k =tan(((kπ)/n)) \k=(0,1,2,3,...,n−1) p(x)=2i(√((1+x^2 )^n ))Π_(k=0) ^(n−1) (x−tan(((kπ)/n)))$
	$p (x) = \sqrt{{(1 + x^{2})}^{n}} (e^{i n θ} - e^{- i n θ}) ∖ θ = {t a n}^{- 1} x$ $= \sqrt{{(1 + x^{2})}^{n}} (2 i s i n (n θ))$ $p (x) = 2 i \sqrt{{(1 + x^{2})}^{n}} s i n ({n t a n}^{- 1} (x))$ ${t a n}^{- 1} (x_{k}) = \frac{k π}{n} \Rightarrow x_{k} = t a n (\frac{k π}{n})$ $s o r o o t s o f p (x) a r e α_{k} = t a n (\frac{k π}{n}) ∖ k = (0, 1, 2, 3, . . ., n - 1)$ $p (x) = 2 i \sqrt{{(1 + x^{2})}^{n}} \prod_{k = 0}^{n - 1} (x - t a n (\frac{k π}{n}))$
	Commented by abdo imad last updated on 24/Feb/18

	$y o u h a v e c o m m i t e d a e r r o r i n l e a d i n g c o e f f i c i e n t o f p (x)$ $s i r .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum