How to do that complex number question

u+v =
w+ w^2 + w^3 + w^4 + 1 - 1 =
(w-1)(1+w+ w^2 + w^3 + w^4)/(w-1) -1 = (w^5-1)/(w-1) -1 = (1-1)/( e^(2I(pi)/5) -1) - 1 = -1

uv =
(w+w^4)(w^2+w^3) = w^3 + w^4 + w^6 +w^7 =
w^3 + w^4 + w*w^5 +w^2*w^5 = w+ w^2 +w^3+w^4 = - 1

(u - v)^2 = (u+v)^2 - 4uv = (-1)^2 - 4(-1) = 5

Exact value of cos(2π/5)
Using Euler
( cos(2π/5) + i sin(2π/5))^5 = cos(2π) + i sin(2π) = 1

Using Newton
( cos(2π/5) + i sin(2π/5))^5 =
cos^5 (2π/5) + 5icos^4(2π/5)sin(2π/5) - 10cos^3(2π/5) sin^2(2π/5) - 10icos^2(2π/5) sin^3(2π/5) +
5cos(2π/5) sin^4(2π/5) + i sin^5(2π/5)
Then
cos^5 (2π/5) - 10cos^3(2π/5) sin^2(2π/5) +5cos(2π/5) sin^4(2π/5) = 1
cos^5 (2π/5) - 10cos^3(2π/5) (1- cos^2(2π/5)) +5cos(2π/5)(1- cos^2(2π/5))^2 = 1
cos^5 (2π/5) - 10cos^3(2π/5) + 10cos^5(2π/5) +5cos(2π/5)(1+cos^4(2π/5)-2cos^2(2π/5)) = 1
cos^5 (2π/5) - 10cos^3(2π/5) + 10cos^5(2π/5) +5cos(2π/5) +5cos^5(2π/5)-10cos^3(2π/5) = 1
16cos^5 (2π/5) - 20cos^3(2π/5) +5cos(2π/5) - 1 = 0
(cos(2π/5) -1 )(16cos^4(2π/5)+ 16cos^3(2π/5) -4cos^2(2π/5) -4cos(2π/5)+1) = 0
16cos^4(2π/5)+ 16cos^3(2π/5) -4cos^2(2π/5) -4cos(2π/5)+1 = 0
(4cos^2(2π/5)+ 2cos(2π/5))^2 - 2 [4cos^2(2π/5) + 2cos(2π/5)] + 1 = 0
(4cos^2(2π/5)+ 2cos(2π/5) -1)^2 = 0
4cos^2(2π/5)+ 2cos(2π/5) -1 = 0
cos(2π/5) = (-1 ± √5)/4
Right is
cos(2π/5) = ( √5 - 1)/4