[1]n=2
[1,cosπ/2]=[1,0]
[0,sinπ/2] [0,1]
(2/2)^1/2をかける.
[2]n=3
[1,cos2π/3,cos4π/3] [1,-1/2,-1/2]
[0,sin2π/3,sin4π/3]=[0,√3/2,-√3/2]
[1/√2,1/√2,1/√2] [1/√2,1/√2,1/√2]
(2/3)^1/2をかける.
[√(2/3),-√(1/6),-√(1/6)]
=[0,1/√2,-1/√2]
[1/√3,1/√3,1/√3]
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