Find-Z-4-1-Hence-show-that-1-w-w-2-w-3-0-

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 \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} . .

★ \dots 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