ilovewolfoo

show your work and no image

Q: 

let w = e^(4i*pi/7). Evaluate (2+w)(2+w^2)(2+w^3)(2+w^4)(2+w^5)(2+w^6)

Feel free to type your answer in the comments box. Does this look familliar to you?

A: We encounter the product of binomials in the seventh roots of unity. 

\(z^7-1 = (z-1)(z-r_1)(z-r_2)(z-r_3)(z-r_4)(z-r_5)(z-r_6), \) where all of the \(r_i\) are the 7th roots of unity besides 1. We recognize that \(w\) is a 7th root of unity because w^7 = 1. 

so  \(z^7-1 = (z-1)(z-w)(z-w^2)(z-w^3)(z-w^4)(z-w^5)(z-w^6).\) We can let z = 2 into the equation and get

-129 =  -3 (-1)^6 (2+w)(2+w^2)(2+w^3)(2+w^4)(2+w^5)(2+w^6) the answer is 43

That's a very good question! The area of the polygon is 14. I actually did not understand this problem at first, but then I found a way.

Solution: We are looking for positive integers (x, y) such that x^2 + y^2 = 5. We find the solutions (x,y) =

(1,2), (1, -2), (-1, 2), (-1, -2), (2, 1), (2, -1), (-2, 1), (-2, -1), which forms a convex octagon.

To calculate the area of the octagon, you can subtract the area of the right triangles from the area of the square with vertices of (+-2, +-2), where +-a represents a and -a. The area of the square is 16, while the distance between an octagon vertice and a vertice of the square is 1, so the area of each of the triangles is 1/2, so 4 of the triangles has area 2.

The answer is 16 - 2 = 14.

WARNING: Do not assume the octagon was regular! It was equiangular, with 4 sides of length sqrt2 and 4 alternating sides of length 2.

I hope you enjoy the solution! Thank you for reading it!