let p(x) =x^n −e^(inα) with n integr and α fromR 1) find the roots of p(x) 2) factorize p(x) inside C[x] .


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Question Number 36352 by prof Abdo imad last updated on 01/Jun/18

$l e t p (x) = x^{n} - e^{i n α} w i t h n i n t e g r a n d α f r o m R$ $1) f i n d t h e r o o t s o f p (x)$ $2) f a c t o r i z e p (x) i n s i d e C [x] .$
Commented by math khazana by abdo last updated on 10/Jul/18
$changement x =e^(iα ) z give p(x)=0 ⇔ e^(inα) z^n −e^(inα) =0 ⇔z^n −1 =0 ⇔ z_k =e^((i2kπ)/n) and k∈[[0,n−1]] so the roots of p(x) are x_k =e^(iα) e^(i((2kπ)/n)) =e^(i(((2kπ)/n) +α)) with 0≤k≤n−1 2) wehave p(x) = Π_(k=0) ^(n−1) {x− e^(i(((2kπ)/n) +α)) }$
$c h a n g e m e n t x = e^{i α} z g i v e$ $p (x) = 0 \Leftrightarrow e^{i n α} z^{n} - e^{i n α} = 0 \Leftrightarrow z^{n} - 1 = 0 \Leftrightarrow$ $z_{k} = e^{\frac{i 2 k π}{n}} a n d k \in [[0, n - 1]] s o t h e r o o t s o f p (x)$ $a r e x_{k} = e^{i α} e^{i \frac{2 k π}{n}} = e^{i (\frac{2 k π}{n} + α)} w i t h 0 ⩽ k ⩽ n - 1$ $2) w e h a v e p (x) = \prod_{k = 0}^{n - 1} {x - e^{i (\frac{2 k π}{n} + α)}}$
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