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Question Number 31496 by abdo imad last updated on 09/Mar/18

l e t a \in] 0, π [a n d A (x) = x^{2 n} - 2 c o s (n a) x^{n} + 1

1) f a c t o r i z e i n s i d e C [x] A (x)

2) f a c t o r i z e i n s i d e R [x] A (x) .

Commented by abdo imad last updated on 14/Mar/18

f i r s t l e t p u t x^{n} = z \Leftrightarrow x =^{n} \sqrt{z} \Rightarrow A (x) = z^{2} - 2 c o s (n a) z + 1

Δ^{^{'}} = {c o s}^{2} (n a) - 1 = - {s i n}^{2} (n a) = {(i s i n (n a))}^{2}

z_{1} = c o s (n a) + i s i n (n a) = e^{i n a} a n d z_{2} = e^{- i n a} \Rightarrow

x = z^{\frac{1}{n}} \Rightarrow x = e^{i a} o r x = e^{- a} \Rightarrow

A (x) = (z - e^{i n a}) (z - e^{- i n a}) = (x^{n} - e^{i n a}) (x^{n} - e^{- i n a})

A (x) = 0 \Leftrightarrow x^{n} = e^{i n a} o r x^{n} = e^{- i n a} \Leftrightarrow {(x e^{- i a})}^{n} = 1 x

o r {(x e^{i a})}^{n} = 1 \Leftrightarrow x e^{- i a} = e^{i \frac{2 k π}{n}} o r x e^{i a} = e^{i \frac{2 k π}{n}} \Rightarrow

x = e^{i (\frac{2 k π}{n} + a)} o r x = e^{i (\frac{2 k π}{n} - a)} s o t h e r o o t s o f A (x) a r e

x_{k} = e^{i (\frac{2 k π}{n} + a)} a n d t_{k} = e^{i (\frac{2 k π}{n} - a)} a n d k \in [[0, n - 1]] . a n d

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

A (x) = \prod_{k = 0}^{n - 1} (x - e^{i (\frac{2 k π}{n} + a)}) \prod_{k = 0}^{n - 1} (x - e^{i (\frac{2 k π}{n} - a)}) b e c o n t i n u e d \dots