Complex Roots of Unit (1 Viewer)

[tharkaan]

Hi,

Just a question regarding complex roots of unity,
(Terry Lee 2.10 –boredofstudies.orgq 2 iv)

The question pretty much says;

if (a)(b) = 3, and a + b = -3

find a and b,

Then you use a quadratic, x^2 + 3x + 3 = 0
and then you get x=+/-....
How do you know which answer belongs to a, and which one belongs to b?

Thanks

 

[ohexploitable]

(a)(b)=(b)(a)

a+b=b+a

a and b can be either

 

[Trebla]

It doesn't matter. Due to the symmetry of the expressions, they are interchangable.

 

[hscishard]

Hi,

Just a question regarding complex roots of unity,
(Terry Lee 2.10 –boredofstudies.orgq 2 iv)

The question pretty much says;

if (a)(b) = 3, and a + b = -3

find a and b,

Then you use a quadratic, x^2 + 3x + 3 = 0
and then you get x=+/-....
How do you know which answer belongs to a, and which one belongs to b?

Thanks

Incorrect

I also want an explanation of Lee's answer. I don't get it when he writes Im[w]=-Im[w^2]....etc

w^3=1
Prove that (1+2w+3w^2)(1+3w+2e^2)=3 and (1+2w+3w^2) + (1+3w+w^2)=-3, hence find the exact values of (1+2w+3w^2) and (1+3w+w^2)

 

[deterministic]

Suppose and
Then
Thus


Using the fact gives

Do something similar to the other expression. Then solving the relevant quadratic, we can see that the difference will lie in the imaginary bit (roots are conjugates and hence only the imaginary bits differ).

Thus using the first fact I shown:

and


BUT since we don't specify which complex root of unity w is (we just note it is not 1), then as pointed out above, both expression can take either one of the solutions of the quadratic.