let p(x)=(1+jx)^n −(1−jx)^n with j=e^((i2π)/3) 1) determine the roots of p(x) and factorize P(x) inside C[x] 2) decompose the fraction F(x)=(1/(p(x)))


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Question Number 74225 by mathmax by abdo last updated on 20/Nov/19
$let p(x)=(1+jx)^n −(1−jx)^n with j=e^((i2π)/3) 1) determine the roots of p(x) and factorize P(x) inside C[x] 2) decompose the fraction F(x)=(1/(p(x)))$
$l e t p (x) = {(1 + j x)}^{n} - {(1 - j x)}^{n} w i t h j = e^{\frac{i 2 π}{3}}$ $1) d e t e r m i n e t h e r o o t s o f p (x) a n d f a c t o r i z e P (x) i n s i d e C [x]$ $2) d e c o m p o s e t h e f r a c t i o n F (x) = \frac{1}{p (x)}$
	Answered by mind is power last updated on 20/Nov/19
	$p(x)=0⇒(1+jx)^n −(1−jx)^n =0 ⇔(((1+jx)/(1−jx)))^n =1 &x#(1/j) ((1+jx)/(1−jx))=e^((2ikπ)/n) ,x≠(1/j) k<n ⇔x=((e^((2ikπ)/n) −1)/(j(e^((2ikπ)/n) +1)))=(1/j) ((e^(i((kπ)/n)) −e^((−ikπ)/n) )/(e^(ik(π/n)) +e^(−((ikπ)/n)) ))=((i tan(((kπ)/n)))/j) x=(i/j).tan(((kπ)/(2n))) k∈[0,n−1] withe (k/n)≠2 if n=2s⇒k∈[1,2s−1]−{s} if n=2s+1⇒k∈[1,2s−1] p(x)=(1+jx)^n −(1−jx)^n p(x)=(jx)^n −(−jx)^n +n(jx)^(n−1) −n(−jx)^(n−1) ........ p(x)=(j^n +(−1)^(n+1) j^j )x^n +n(j^(n−1) +(−1)^n j^(n−1) )x^(n−1) if n=2s p is polynom degp=2s−1 if n=2s+1 degp=2s+1 n=2s p(x)= n(2j^(n−1) )Π_(k=0,k≠(n/2)) ^(n−1) (X−((itan(((kπ)/n)))/j)) if n=2s+1 p(x)=(2j^n )Π_(k=0) ^(n−1) (X−((itan(((kπ)/n)))/j)) 2)(1/(p(x))) we do it for n=2k+1 (1/(p(x)))=Σ_(k=0) ^(n−1) (a_k /((X−((itan(((kπ)/n)))/j)))) a_k =(1/(2j^n Π_(t=0,t#k) ^(n−1) (((i(tan(((tπ)/n))−tan(((kπ)/n)))/j)))) (1/(p(x)))=Σ_(k=0) ^(n−1) (1/(2j^n Π_(t=0,t#k) ^(n−1) (((i(tan(((tπ)/n))−tan(((kπ)/n)))/j)))).(1/((X−((itan(((kπ)/n)))/j)))) .$
	$p (x) = 0 \Rightarrow {(1 + j x)}^{n} - {(1 - j x)}^{n} = 0$ $You can't use 'macro parameter character #' in math mode$ $\frac{1 + j x}{1 - j x} = e^{\frac{2 i k π}{n}}, x \neq \frac{1}{j} k < n$ $\Leftrightarrow x = \frac{e^{\frac{2 i k π}{n}} - 1}{j (e^{\frac{2 i k π}{n}} + 1)} = \frac{1}{j} \frac{e^{i \frac{k π}{n}} - e^{\frac{- i k π}{n}}}{e^{i k \frac{π}{n}} + e^{- \frac{i k π}{n}}} = \frac{i t a n (\frac{k π}{n})}{j}$ $x = \frac{i}{j} . t a n (\frac{k π}{2 n}) k \in [0, n - 1] w i t h e \frac{k}{n} \neq 2$ $i f n = 2 s \Rightarrow k \in [1, 2 s - 1] - {s}$ $i f n = 2 s + 1 \Rightarrow k \in [1, 2 s - 1]$ $p (x) = {(1 + j x)}^{n} - {(1 - j x)}^{n}$ $p (x) = {(j x)}^{n} - {(- j x)}^{n} + n {(j x)}^{n - 1} - n {(- j x)}^{n - 1} . . . . . . . .$ $p (x) = (j^{n} + {(- 1)}^{n + 1} j^{j}) x^{n} + n (j^{n - 1} + {(- 1)}^{n} j^{n - 1}) x^{n - 1}$ $i f n = 2 s p i s p o l y n o m d e g p = 2 s - 1$ $i f n = 2 s + 1 d e g p = 2 s + 1$ $n = 2 s$ $p (x) = n (2 j^{n - 1}) \underset{k = 0, k \neq \frac{n}{2}}{\prod^{n - 1}} (X - \frac{i t a n (\frac{k π}{n})}{j})$ $i f n = 2 s + 1$ $p (x) = (2 j^{n}) \underset{k = 0}{\prod^{n - 1}} (X - \frac{i t a n (\frac{k π}{n})}{j})$ $2) \frac{1}{p (x)}$ $w e d o i t f o r n = 2 k + 1$ $\frac{1}{p (x)} = \underset{k = 0}{\sum^{n - 1}} \frac{a_{k}}{(X - \frac{i t a n (\frac{k π}{n})}{j})}$ $You can't use 'macro parameter character #' in math mode$ $You can't use 'macro parameter character #' in math mode$ $.$
	Commented by mathmax by abdo last updated on 21/Nov/19

	$t h a n k x s i r .$
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