# Questions about comlex number --- complex root of unity

Suppose that w is the complex root of unity (w =/= 1).

Show that

(a) (1 - w + w^2)(1 + w -w^2) = 4

(b) (2 + 5w +2w^2)^6 = (2 + 2w + 5w^2)^6 = 729

### 2 個解答

• 最愛解答

w^3 = 1............(1)

1 + w + w^2 = 0.............(2)

(a)

(1 - w + w^2)(1 + w - w^2) = [(1 + w + w^2) - 2w][( 1 + w + w^2) - 2w^2]

= ( 0 - 2w)(0 - 2w^2) = 4w^3 = 4.

(b)

(2 + 5w + 2w^2)^6 = [2(1 + w + w^2) + 3w]^6 = ( 2 x 0 + 3w)^6

= (3w)^6 = 729(w^3)^2 = 729.

(2 + 2w + 5w^2)^6 = [(2(1 + w + w^2) + 3w^2]^6 = ( 2 x 0 + 3w^2)^6

= (3w^2)^6 = 729w^12 = 729(w^3)4 = 729.

Therefore, (2 + 5w + 2w^2)^6 = (2 + 2w + 5w^2)^6 = 729. Your question is correct.

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• You have some information missed in your question.

w should be the complex cubic roots of unity (w =/= 1)

So, w3 = 1

1 + w + w2 = 0

a. (1 - w + w2)(1 + w - w2)

= [1 - w(1 - w)][1 + w(1 - w)]

= 1 - w2(1 - w)2

= 1 - w2 + 2w3 - w4

= (1 + 2w3) - (w2 + w4)

= 1 + 2 - [w2 + w3‧w]

= 3 - (w2 + w)

= 3 - (w2 + w + 1) + 1

= 3 - 0 + 1

= 4

b. There is some typing errors in your question, it should be:

(2 + 5w + 2w2)6 = (5 + 2w + 5w2)6

(2 + 5w + 2w2)6

= [2 + 5(-w2 - 1) + 2(-w - 1)]6

= (-5 - 2w - 5w2)6

= (-1)6(5 + 2w + 5w2)6

= (5 + 2w + 5w2)6

(5 + 2w + 5w2)6

= [5(1 + w + w2) - 3w]6

= [5(0) - 3w]6

= (-3)6w6

= 729(w3)2

= 729(1)2

= 729

Hence, we have shown (2 + 5w + 2w2)6 = (5 + 2w + 5w2)6 = 729

2008-09-14 12:20:40 補充：

Now, (2 + 5w + 2w^2)^6 = (2 + 2w + 5w^2)^6 = 729

Because (2 + 2w + 5w^2)^6

= [2(1 + w + w^2) + 3w^2]^6

= [2(0) + 3w^2]^6

= (3^6)w^12

= 729(w^3)^4

= 729(1)^4

= 729

So, we have shown (2 + 5w + 2w^2)^6 = (2 + 2w + 5w^2)^6 = 729

2008-09-14 12:22:07 補充：

So, there is no typing error in your question b. Sorry for the tiny mistake I have made.

Anyways, your question should state that w is a complex cubic root of unity.

資料來源： Myself~~~
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