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Question Number 49647 by maxmathsup by imad last updated on 08/Dec/18

l e t p (x) = x^{2 n} - x^{n} + 1

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

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

3) s o l v e p (x) = 0 a n d p (x) = 2

Commented by maxmathsup by imad last updated on 08/Dec/18

l e t p u t x^{n} = t \Rightarrow x^{2 n} - x^{n} + 1 = t^{2} - t + 1

Δ = 1 - 4 = - 3 = {(i \sqrt{3})}^{2} \Rightarrow t_{1} = \frac{1 + i \sqrt{3}}{2} a n d t_{2} = \frac{1 - i \sqrt{3}}{2}

\Rightarrow t_{1} = e^{\frac{i π}{3}} a n d t_{2} = e^{- \frac{i π}{3}} \Rightarrow p (x) = (x^{n} - e^{\frac{i π}{3}}) (x^{n} - e^{- \frac{i π}{3}}) s o p (x) = 0 \Leftrightarrow x^{n} = e^{\frac{i π}{3}}

o r x^{n} = e^{- \frac{i π}{3}} l e t s o l v e x^{n} = e^{\frac{i π}{3}} x = r e^{i θ} \Rightarrow r^{n} e^{i n θ} = e^{\frac{i π}{3}} \Rightarrow r = 1 a n d n θ = \frac{π}{3} + 2 k π \Rightarrow

θ_{k} = \frac{π}{3 n} + \frac{2 k π}{n} w i t h k \in [[0, n - 1]] \Rightarrow x_{k} = e^{i (\frac{π}{3 n} + \frac{2 k π}{n})}

a l s o x^{n} = e^{- \frac{i π}{3}} g i v e t_{k} = e^{i (- \frac{π}{3 n} + \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

x_{k} a n d t_{k} .

2) p (x) = \prod_{k = 0}^{n - 1} (x - x_{k}) \prod_{k = 0}^{n - 1} (x - t_{k})

= \prod_{k = 0}^{n - 1} (x - x_{k}) (x - {\bar{x}}_{k}) w i t h x_{k} = e^{i (\frac{π}{3 n} + \frac{2 k π}{n})}

Commented by maxmathsup by imad last updated on 08/Dec/18

3) p (x) = 0 \Leftrightarrow x = x_{k} o r x = t_{k} \Leftrightarrow x = e^{i (\frac{π}{3 n} + \frac{2 k π}{n}}) o r x = e^{i (- \frac{π}{3 n} + \frac{2 k π}{n})}

k \in {0, 1, 2, \dots, n - 1}

Commented by maxmathsup by imad last updated on 08/Dec/18

p (x) = 2 \Rightarrow x^{2 n} - x^{n} + 1 = 2 \Rightarrow x^{2 n} - x^{n} - 1 = 0

Δ = 1 + 4 = 5 \Rightarrow x^{n} = \frac{1 + \sqrt{5}}{2} o r x^{n} = \frac{1 - \sqrt{5}}{2}

x^{n} = a (a = \frac{1 + \sqrt{5}}{2}) \Rightarrow x^{n} = {(^{n} \sqrt{a})}^{n} \Rightarrow {(\frac{x}{(^{n} \sqrt{a})})}^{n} = e^{i (2 k π)} \Rightarrow

\frac{x}{(^{n} \sqrt{a})} = e^{\frac{i 2 k π}{n}} \Rightarrow x_{k} =^{n} \sqrt{a} e^{\frac{i 2 k π}{n}} w i t h k f r o m [[0, n - 1]]

a l s o x^{n} = \frac{1 - \sqrt{5}}{2} (= α) \Rightarrow x =^{n} \sqrt{α} e^{\frac{i 2 k π}{n}} w i t h 0 ⩽ k ⩽ n - 1

Answered by Smail last updated on 08/Dec/18

1)

x^{2 n} - x^{n} + 1 = {(x^{n})}^{2} - x^{n} + 1 = (x^{n} + j) (x^{n} + j^{2})

w i t h j = e^{i \frac{2 π}{3}}

x^{n} = - j = e^{i \frac{5 π}{3}} o r x^{n} = - j^{2} = e^{i \frac{π}{3}}

x = e^{i \frac{π}{3 n} (5 + 6 k)} o r x = e^{i \frac{π}{3 n} (6 k + 1)} w i t h k = (0, 1, 2, \dots, n - 1)

2)

p (x) = \underset{k = 0}{\prod^{n - 1}} (x - e^{i \frac{π}{3 n} (6 k + 5)}) (x - e^{i \frac{π}{3 n} (6 k + 1)})

3)

p (x) = 0 w h e n x = e^{i \frac{π}{3 n} (6 k + 5)} o r x = e^{i \frac{π}{3 n} (6 k + 1)}

p (x) = 2 \Leftrightarrow x^{2 n} - x^{n} + 1 = 2

x^{2 n} - x^{n} - 1 = 0

Δ = 1 + 4 = 5

x^{n} = \frac{1 \underline{+} \sqrt{5}}{2} = a

x = \sqrt[n]{a} e^{\frac{2 i k π}{n}} = \sqrt[n]{\frac{1 \underline{+} \sqrt{5}}{2}} e^{\frac{2 i k π}{n}}

Commented by afachri last updated on 08/Dec/18

pardon me Mr . Smail, may i ask you a question

? cause i am a little bit confused, i am

a new beginner . question :

can we write down x^{n} to

x^{n} = \frac{1 \pm i \sqrt{3^{}}}{2} ? and would you mind

explain to me why j = e^{i \frac{2 π}{3}} ? ?

Thank You, Sir

Commented by Smail last updated on 08/Dec/18

O f c o u r s e y o u c a n .

j = e^{i \frac{2 π}{3}} i s j u s t a s i m p l e w a y t o w r i t e t h e s o l u t i o n .

\frac{1 + i \sqrt{3}}{2} = \frac{1}{2} + i \frac{\sqrt{3}}{2} = c o s (\frac{π}{3}) + i s i n (\frac{π}{3})

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

= - j

w i t h j^{3} = 1

Commented by Smail last updated on 08/Dec/18

A l s o j^{2} + j + 1 = 0

Commented by afachri last updated on 09/Dec/18

thank you sir, i got it .

thank you very much :)

Commented by Smail last updated on 09/Dec/18

y o u a r e w e l c o m e

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