Find Z^4 =1. Hence show that 1+w+w^2 +w^3 =0


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Question Number 155572 by peter frank last updated on 02/Oct/21

$Find Z^{4} = 1 .$ $Hence show that 1 + w + w^{2} + w^{3} = 0$
	Answered by puissant last updated on 02/Oct/21
	$z^4 =1 ⇒ z=e^(i((2kπ)/4)) ; k∈[∣0;3∣].. → k=0 ⇒ z=1.. → k=1 ⇒ z=e^(i(π/2)) = i.. → k=2 ⇒ z=e^(i((4π)/4)) = e^(iπ) =−1.. → k=3 ⇒z=e^(i((3π)/2)) = −i.. ∵ z ∈ {−1 ; −i ; i ; 1 }.. ★... 1+w+w^2 +w^3 =((1−w^4 )/(1−w))= ((w^4 −1)/(w−1)) w is a solution, we have w^4 = 1 ⇒ w^4 −1=0 ⇒((w^4 −1)/(w−1))=0 ⇒ 1+w+w^2 +w^3 =0..$
	$z^{4} = 1 \Rightarrow z = e^{i \frac{2 k π}{4}}; k \in [∣ 0; 3 ∣] . .$ $\to k = 0 \Rightarrow z = 1 . .$ $\to k = 1 \Rightarrow z = e^{i \frac{π}{2}} = i . .$ $\to k = 2 \Rightarrow z = e^{i \frac{4 π}{4}} = e^{i π} = - 1 . .$ $\to k = 3 \Rightarrow z = e^{i \frac{3 π}{2}} = - i . .$ $∵ z \in {- 1; - i; i; 1} . .$ $★ . . . 1 + w + w^{2} + w^{3} = \frac{1 - w^{4}}{1 - w} = \frac{w^{4} - 1}{w - 1}$ $w i s a s o l u t i o n, w e h a v e w^{4} = 1$ $\Rightarrow w^{4} - 1 = 0$ $\Rightarrow \frac{w^{4} - 1}{w - 1} = 0 \Rightarrow 1 + w + w^{2} + w^{3} = 0 . .$
	Commented by peter frank last updated on 02/Oct/21

	$great sir . thanks$
	Answered by Rasheed.Sindhi last updated on 02/Oct/21

	$z^{4} - 1 = 0$ $(z - 1) (z + 1) (z^{2} + 1) = 0$ $z = 1, z = - 1, z = \pm i$ $ω = i (s a y)$ $ω^{2} = i^{2} = - 1$ $ω^{3} = i^{3} = i^{2} . i = - i$ $ω^{4} = i^{4} = i^{2} . i^{2} = - 1 . - 1 = 1$ $1 + ω + ω^{2} + ω^{3} = 1 + i + (- 1) + (- i) = 0$
	Commented by peter frank last updated on 02/Oct/21

	$thank you$
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