18/01/21 01:59:37.38 SUkk+U6n.net
>>41 (蛇足）
>>51 より
ζ = 1 - ｅ^(iπ/4）= -2ｉ sin(π/8）ｅ^(iπ/8),
ζ^2 = -2{1 - cos(π/4)}ｅ^(iπ/4）= -(2-√2）ｅ^(iπ/4),
ζ^4 = 2(√2 -1)^2 i
ζ^8 = -4(√2 -1)^4 ∈ Q[√2］⊂ R
18/01/21 01:59:37.38 SUkk+U6n.net
>>41 (蛇足）
>>51 より
ζ = 1 - ｅ^(iπ/4）= -2ｉ sin(π/8）ｅ^(iπ/8),
ζ^2 = -2{1 - cos(π/4)}ｅ^(iπ/4）= -(2-√2）ｅ^(iπ/4),
ζ^4 = 2(√2 -1)^2 i
ζ^8 = -4(√2 -1)^4 ∈ Q[√2］⊂ R