# roots of unity questions (1 Viewer)

#### poptarts12345

##### Member
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

##### Well-Known Member
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

##### Member
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: $\bg_white w^n=1$ by definition. dividing by w on both sides: $\bg_white w^{n-1}=\frac{1}{w}$ Also note that $\bg_white |w|=1$ so $\bg_white w^{n-1}=\frac{\overline{w}}{w\overline{w}}=\overline{w}$ since $\bg_white w\overline{w}=|w|^2$

#### idkkdi

##### Well-Known Member
For 2: $\bg_white w^n=1$ by definition. dividing by w on both sides: $\bg_white w^{n-1}=\frac{1}{w}$ Also note that $\bg_white |w|=1$ so $\bg_white w^{n-1}=\frac{\overline{w}}{w\overline{w}}=\overline{w}$ since $\bg_white w\overline{w}=|w|^2$
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

##### Member
clearly w+w^2+w^4 + w^3+w^5+w^6 = -1.
How did you come to this conclusion?

#### idkkdi

##### Well-Known Member
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

##### Member
How did you come to this conclusion?
If you want to know how he came up with that factorisation: Notice that $\bg_white 1+w+w^2+...+w^6$ is a GP so using the GP sum: $\bg_white 1+w+w^2+...+w^6=\frac{w^7-1}{w-1}$ now we know $\bg_white w^7=1$ and $\bg_white w \neq 1$ so $\bg_white 1+w+w^2+...+w^6=0$ subtract 1 from both sides you get $\bg_white w+w^2+...+w^6=-1$

#### CM_Tutor

##### Well-Known Member
2. Let w be the principal nth root of unity.
Prove that w conjugate = w^(n-1)
Another (similar) approach to this part is:

\bg_white \begin{align*} \bar{w} &= \bar{w} \times \frac{w}{w} \\ &= \frac{w\bar{w}}{w} \\ &= \frac{|w|^2}{w} \qquad \text{as w\bar{w} = |w|^2 for any w \in \mathbb{C}} \\ &=\frac{1}{w} \qquad \text{as w is a root of unity and so |w|=1} \\ &= \frac{w^n}{w} \qquad \text{as w is an nth root of unity and so w^n=1} \\ &= w^{n-1} \qquad \text{as required} \end{align*}