%3Cdiv%20id%3D%22MathJax_Message%22%26nbsp%3Bstyle%3D%22display%3A%26nbsp%3Bnone

Question

%3Cdiv%20id%3D%22MathJax_Message%22%26nbsp%3Bstyle%3D%22display%3A%26nbsp%3Bnone%3B%22%26nbsp%3Bdata-mce-style%3D%22display%3A%26nbsp%3Bnone%3B%22%3E%3C%2Fdiv%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%3C%2Fp%3E%3Cp%3E%3Cimg%26nbsp%3Bclass%3D%22%22%26nbsp%3Bsrc%3D%22http%3A%2F%2Fmedia.cheggcdn.com%2Fmedia%252F62d%252F62d3d207-8c18-4d34-87f8-bb3b294f306e%252Fphpp2vYDJ.png%22%26nbsp%3Bheight%3D%22600%22%26nbsp%3Bwidth%3D%221007%22%26nbsp%3Bdata-mce-src%3D%22http%3A%2F%2Fmedia.cheggcdn.com%2Fmedia%252F62d%252F62d3d207-8c18-4d34-87f8-bb3b294f306e%252Fphpp2vYDJ.png%22%3E%3C%2Fp%3E

Explanation / Answer

(the sum of wn^ri , r = 0 to n-1) * (1 - wn^i) = 1 - wn ^ ni   (this is a telescoping sum)

By definition, wn ^ n = 1, and wn ^ ni = 1, so 1 - wn ^ ni = 0

Thus, one of the factors, (the sum of wn ^ri , r = 0 to n-1) or (1 - wn^i) must equal 0.

wn^i does not equal 1. Thus, (the sum of wn ^ri , r = 0 to n-1) = 0

The product of wn^r, r=0 to n-1 =,   summing the exponents, wn ^ [(n)(n-1)]/2

Consider n even and n odd

If n is even, as wn^n= 1, and n/2 is an integer, wn^(n/2)^2 = 1, so wn^(n/2)= plus or - 1. However, if wn^(n/2) = 1, we will only have at most n/2 roots. Thus, wn^(n/2) = -1.

Then, as n-1 is odd, wn^(n/2)^(n-1) = (-1)^(n-1) = -1. Thus, wn ^ [(n)(n-1)]/2 = -1, and

The product of wn^r, r=0 to n-1 = -1

Next, consider n odd.

Then, wn^[(n)(n-1)]/2 = , as (n-1)/2 is an interger, (wn^n)^(n-1)/2 = 1^(n-1)/2 = 1

The product of wn^r, r=0 to n-1 = 1

Thus, for n even or odd, The product of wn^r, r=0 to n-1 = (-1)^(n-1)

b)

Factor the polynomial as (x + w3y + w3^2 z)(x + w3^2y + w3 z) =

x^2 + w3^3y^2 + w3^3z^2 + xy(w3 + w3^2) + xz(w3 + w3^2) + yz(w3^2 + w3^4)

w3^4 = w3^3*w3 = 1*w3 = w3 w3^3 = 1

Thus, x^2 + w3^3y + w3^3z + xy(w3 + w3^2) + xz(w3 + w3^2) + yz(w3^2 + w3) =

x^2 + y^2 + z^2 + xy(w3 + w3^2) + xz(w3 + w3^2) + yz(w3 + w3^2)

From ai) 1 + w3 + w3^2 = 0. Thus, w3 + w3^2 = -1. Thus,

x^2 + y^2 + z^2 + xy(w3 + w3^2) + xz(w3 + w3^2) + yz(w3 + w3^2) =

x^2 + y^2 + z^2 - xy - xz - yz

Note: as we know that w3 = -1/2 + sqrt(3)/2 i, and w3^2 = -1/2 - sqrt(3)/2 i , we may write these factors as

(x + (-1/2+sqrt(3)/2i)y + (-1/2-sqrt(3)/2i)z) (x + (-1/2-sqrt(3)/2i)y + (-1/2+sqrt(3)/2i)z) =

x^2 + y^2 + z^2 -xy - xz - yz

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.