roots of unity questions (1 Viewer)

[poptarts12345]

1. Let w be a seventh root of unity. Find the equation of the quadratic polynomial with roots w+w^2+w^4 and w^3+w^5+w^6.
ANS : x^2+x+2
2. Let w be the principal nth root of unity.
Prove that w conjugate = w^(n-1)

 

[idkkdi]

1. Let w be a seventh root of unity. Find the equation of the quadratic polynomial with roots w+w^2+w^4 and w^3+w^5+w^6.
ANS : x^2+x+2
2. Let w be the principal nth root of unity.
Prove that w conjugate = w^(n-1)

w^7-1 = 0. (w-1)(w^6+w^5+...+1 ) = 0. For w =/ 1, (w^6+w^5+...+1) = 0. clearly w+w^2+w^4 + w^3+w^5+w^6 = -1.
(w+w^2+w^4)(w^3+w^5+w^6) = w^4 + w^6 + w^7 + w^5 + w^7 + w^8 + w^7 + w^9 + w^10 = w^4 + w^5 + w^6 + w^7 + w^7 + w + w^2 + w^3 + w^7 = -1 + 3 = 2.

(note that w is assumed to be the principal root of unity)
let w = cis theta
w cong = cis -theta
w^n-1 = cis (n-1) theta
= cis n theta/cis theta
= cis -theta

 

[Qeru]

1. Let w be a seventh root of unity. Find the equation of the quadratic polynomial with roots w+w^2+w^4 and w^3+w^5+w^6.
ANS : x^2+x+2
2. Let w be the principal nth root of unity.
Prove that w conjugate = w^(n-1)

For 2: by definition. dividing by w on both sides: Also note that so since

 

[idkkdi]

For 2: by definition. dividing by w on both sides: Also note that so since

i was tempted to write by definition as well, but that doesn't match up with the decently long algebra required for the first q lol.

 

[poptarts12345]

clearly w+w^2+w^4 + w^3+w^5+w^6 = -1.

How did you come to this conclusion?

 

[idkkdi]

How did you come to this conclusion?

w^7-1 = 0. (w-1)(w^6+w^5+...+1 ) = 0. For w =/ 1, (w^6+w^5+...+1) = 0. clearly w+w^2+w^4 + w^3+w^5+w^6 = -1.

 

[Qeru]

How did you come to this conclusion?

If you want to know how he came up with that factorisation: Notice that is a GP so using the GP sum: now we know and so subtract 1 from both sides you get

 

[CM_Tutor]

2. Let w be the principal nth root of unity.
Prove that w conjugate = w^(n-1)

Another (similar) approach to this part is: