Math Complex numbers?

x^8 = 1 = cos(2npi) + isin(2npi)

x= (cos(2npi) + isin(2npi) )^1/8

= cos(npi/8) + i sin(npi/8) using Demoivre thn

For n=1 we have a =cosπ/4+i sinπ/4

for n = 2 we have cos2π/4+i sinπ2/4 = a^2 ( DeMoivre again)

for n = 3 we have cos3π/4+i sin3π/4 = a^3

So the roots are a, a^2, ...,a^8=1

SInce this is a polynomial the sum of roots is the coefficient of x^7 which is 0

so 1+α+α²+α³+......+α⁷=0

x⁸ = 1 so x⁸ - 1 = 0

so 1, -1, i, and -i are solutions as they are solution of x⁴ = 1, and just as (-1)² = 1, and i² = (-i)² = -1, so too the remaining solutions will square to i and -i. They are √½ + i√½, -√½ + i√½, -√½ - i√½ and √½ - i√½.

α¹ = cos(π/4) + i sin(π/4) = √½ + i√½, and α⁰ = 1+0i, so

α² = (α¹)² = (√½ + i√½)² = 0+i

α³ = α¹α² = (√½ + i√½)(0+i) = -√½ + i√½

α⁴ = (α²)² = (0+i)² = -1+0i

α⁵ = α¹α⁴ = (√½ + i√½)(-1+0i) = -√½ - i√½

α⁶ = (α²)³ = (0+i)³ = 0-i

α⁷ = α¹α⁶ = (√½ + i√½)(0-i) = √½ - i√½

This accounts for all our solutions above.

It follows by inspection that α⁰ = -α⁴, α¹ = -α⁵, α² = -α⁶, and α³ = -α⁷, so

1 + α + α² + α³ + α⁴ + α⁵ + α⁶ + α⁷ = α⁰ + α¹ + α² + α³ - α⁰ - α¹ - α² - α³ = 0

Argand diagram: https://www.desmos.com/calculator/r8k1mfbsjh