tough question (1 Viewer)

[somedude2387]

if w is a complex root of z^3 - 1 =0 where w does not equal to 1
show that 1 + w + w^2 = 0 (easy)
and hence evaluate
(1 - w)(1 - w^2)(1 - w^4)(1 - w^8)

help me out

 

[turtle_2468]

Note that as w^3=1,
expression =(1-w)(1-w^2)(1-w)(1-w^2)
=((1-w)(1-w^2))^2
=(w^3-w^2-w+1)^2
=(w^3+2)^2 (as -w^2-w=1 from part 1)
=(1+2)^2
=9

 

[CM_Tutor]

This sort of question pops up for tests / half yealies in this sort of complex roots of unity stuff. The golden rule is to always re-write powers higher than those in the original equation - in this case wⁿ where n > 3 - in terms of lower powers.

As a further practice on this, what is the value of (1 + w)(1 + w²)(1 + w⁴)(1 + w⁸) ?

 

[Heinz]

Originally posted by CM_Tutor

As a further practice on this, what is the value of (1 + w)(1 + w²)(1 + w⁴)(1 + w⁸) ?

(1 + w)(1 + w^2)(1 + w^4)(1 + w^8)
= (1 + w)(1 + w^2)(1 + w)(1 + w^2)
= [(1+w)(1+w^2)]^2
= (1+w+w^2+w^3)^2
= (2 +w + w^2)^2
=(2 - 1)^2
=1

is that right?

 

[CM_Tutor]

Yep, but there's a short cut possible. Look at the terms in your second line, and then think about the result that all these questions start with (ie 1 + w + w² = 0).

 

[:: ck ::]

Originally posted by CM_Tutor
As a further practice on this, what is the value of (1 + w)(1 + w²)(1 + w⁴)(1 + w⁸) ?

(1+w)(1+w²)(1+w⁴)(1+w⁸)
= [(1 + w)(1+w²)]²
= (1 + w² + w + w³)²
= (0 + 1)² (as w³ = 1, 1 + w + w² = 0
= 1

 

[Heinz]

Originally posted by CM_Tutor
Yep, but there's a short cut possible. Look at the terms in your second line, and then think about the result that all these questions start with (ie 1 + w + w² = 0).

Oh yeah, i think i remember doing something like that before just didnt see it this time.

I had this question in a complex numbers test a couple of months ago. Its pretty similar (little more interesting i think)

if w is a complex root of z^3 = 1
i) show that w^2 + w + 1 = 0
ii) prove that b^3 + c^3 = (b+c)(b+cw)((b+cw^2)

 

[CM_Tutor]

Originally posted by Heinz
Oh yeah, i think i remember doing something like that before just didnt see it this time.

Cool - I only mention it because it's always worth trying to save time in 4u - you can sure use it!

For any who don't see what to do, it's:

(1 + w)(1 + w²)(1 + w⁴)(1 + w⁸)
= (1 + w)(1 + w²)(1 + w)(1 + w²), using w³ = 1
= [(-w²)(w)]²
= w⁶
= 1

 

[Grey Council]

sorry, i didn't quite follow what happened here:
(1 + w)(1 + w2)(1 + w4)(1 + w8)
= (1 + w)(1 + w2)(1 + w)(1 + w2), using w3 = 1
how did you go from the frist line to the second line?

 

[Xayma]

w⁴=w³w
w⁸=w³w³w²

 

[Grey Council]

lol, ta mate.
i'm so dumb, nie?