roots of unity question :( (1 Viewer)

[fuckit1991]

hey was wondering if anyone could help me out with this question...god i hate roots of unity anyways, the question goes,

1, w and w^2 are the cube roots of unity. State the values of w^3 and w^2+w+1. Hence, show that (1+w^2)^12=1 and

(1-w)(1-w^2)(1-w^4)(1-w^5)(1-w^7)(1-w^8)=27

thanks

 

[tommykins]

w^3 = 1
1+w+w^2 = 0

(1+w^2)^12 = (1+w^2+w^3-w^3) = (-w^3)^12 = (w^3)^12 = 1^12 = 1.

 

[fuckit1991]

thanks so much tommy...anyone have any ideas on the second part?

 

[Michaelmoo]

Ok for the second part. Substitute w^3 = 1, you get

LHS:

= (1-w)(1-w^2)(1-w)(1-w^2)(1-w)(1-w^2)

= (1-w)^3 x (1-w^2)^3

= [(1-w)(1-w^2)]^3

=[1 - w^2 -w + w^3]^3

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

Now we use the equations proven before (w^3 = 1, 1+w+w^2=0 so w + w^2 = -1) sub these in we get:

= [1- (-1) + 1]^3

= 3^3

= 27

= RHS

Hope that helps.

 

[fuckit1991]

thanks a lot michael...just one question
how did you derive the first line in the LHS

= (1-w)(1-w^2)(1-w)(1-w^2)(1-w)(1-w^2)

 

[Timothy.Siu]

he changed everything larger than w^3 to less,

i.e. w^4=w^3.w=w since w^3=1

w^7=w^3.w^3.w=w

 

[LordPc]

he simplified things down.

w^8 = (w^3)(w^3)(w^2) = (1)(1)(w^2)

you can take out a w^3 and reduce the powers to make things simpler

 

[Michaelmoo]

o sorry bout that u know how say for example u have

(1-w^5), thats the same as

1-(w^3).(w^2) where the dot means times and since w^3 = 1 then its

1-1.(w^2)

=1-w^2

 

[fuckit1991]

thanks a lot guys...u really helped me out

 

[LordPc]

hey was wondering if anyone could help me out with this question...god i hate roots of unity anyways, the question goes,

1, w and w^2 are the cube roots of unity. State the values of w^3 and w^2+w+1. Hence, show that (1+w^2)^12=1 and

(1-w)(1-w^2)(1-w^4)(1-w^5)(1-w^7)(1-w^8)=27

thanks

seriously, I think these questions are probably the easiest in the entire course. its just simplification. they are so easy I dont even think they would put them in the hsc exam.

 

[lyounamu]

seriously, I think these questions are probably the easiest in the entire course. its just simplification. they are so easy I dont even think they would put them in the hsc exam.

lol, harsh

 

[LordPc]

lol, harsh

but is it not true?

I would have loved for a question like this to have appeared in my MX2 exam