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Question Number 69794 by mathmax by abdo last updated on 27/Sep/19

l e t p (x) = {(x + 1)}^{6} - e^{i α} w i t h α r e a l

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]

3) f a c t o r i z e p (x) i n s i d e R [x]

Commented by mathmax by abdo last updated on 01/Oct/19

1) p (x) = 0 \Leftrightarrow {(x + 1)}^{6} = e^{i α} \Leftrightarrow \frac{{(x + 1)}^{6}}{e^{i α}} = 1 \Leftrightarrow \frac{{(x + 1)}^{6}}{{(e^{\frac{i α}{6}})}^{6}} = 1 \Rightarrow

{((x + 1) e^{- \frac{i α}{6}})}^{6} = e^{i 2 k π} \Rightarrow (x + 1) e^{- \frac{i α}{6}} = e^{\frac{i 2 k π}{6}} \Rightarrow

(x + 1) = e^{\frac{i k π}{3}} e^{\frac{i α}{6}} \Rightarrow x + 1 = e^{\frac{i (2 k π + α)}{6}} \Rightarrow x = e^{i (\frac{2 k π + α)}{6}} - 1

s o t h e r o o t s o f p (x) a r e z_{k} = e^{i (\frac{2 k π + α}{6})} - 1 w i t h k \in [[0, 5]]

2) p (x) = a \prod_{k = 0}^{5} (x - z_{k}) w e s e e t h a t a = 1 \Rightarrow

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

Commented by mathmax by abdo last updated on 01/Oct/19

e r r o r o f t y p o p (x) = \prod_{k = 0}^{5} (x + 1 - e^{i (\frac{2 k π + α}{6})})

Answered by mind is power last updated on 29/Sep/19

p (x) = 0 \Rightarrow

{(x + 1)}^{6} = e^{i a} \Leftrightarrow {(x + 1)}^{6} = e^{i (a + 2 k π)}

\Rightarrow x + 1 = e^{i (a + 2 k π) / 6}, k \in [0, 5]

\Rightarrow x_{k} = - 1 + e^{i \frac{a + 2 k π}{6}}

2) p (x) = \underset{k = 0}{\prod^{5}} (x - (- 1 + e^{i \frac{a + 2 k π}{6}}))

p (x) = \underset{k = 0}{\prod^{5}} (x + 1 - e^{i \frac{a + 2 k π}{6}})

o v e r I R n o t p o s s i b l e o n l y [i f e^{i a} \in I R \Rightarrow a = k π

c a u s e p (- 1) = - e^{i a}