give-the-factorization-inside-C-x-for-p-x-x-4-1-i-3-2-

Question Number 29165 by abdo imad last updated on 04/Feb/18

g i v e t h e f a c t o r i z a t i o n i n s i d e C [x] f o r

p (x) = x^{4} - \frac{1 - i \sqrt{3}}{2} .

Commented by abdo imad last updated on 06/Feb/18

l e t f i n d t h e r o o t s l f p (x) p (z) = 0 \Leftrightarrow z^{4} = \frac{1 - i \sqrt{3}}{2} b u t

1 - i \sqrt{3} = 2 (\frac{1}{2} - i \frac{\sqrt{3}}{2}) = 2 e^{- i \frac{π}{3}} l e t p u t z = r e^{i θ}

p (z) = 0 \Leftrightarrow r^{4} = 2 a n d 4 θ = - \frac{π}{3} + 2 k π k \in [[0, 3]]

θ_{k} = - \frac{π}{12} + \frac{k π}{2} s o t h e r o o t s a r e z_{k} =^{4} \sqrt{2} e^{i (- \frac{π}{12} + \frac{k π}{2})} k \in [[0, 3]] i t s

c l e a r t h a t t h e l e a d i n g c o e f f i c i e n t i s 1 s o

p (x) = \prod_{k = 0}^{3} (x - z_{k}) = (x - z_{0}) (x - z_{1}) (x - z_{2}) (x - z_{3}) w i t h

z_{0} =^{4} \sqrt{2} e^{- i \frac{π}{12}}, z_{1} =^{4} \sqrt{2} e^{i \frac{5 π}{12}}, z_{2} =^{4} \sqrt{2} e^{i \frac{11 π}{12}}, z_{3} =^{4} \sqrt{2} e^{i \frac{17 π}{12}} .