Bs Grewal
Question:-if is a complex cube root of unity, prove that 1+⍵+⍵^2 =0
Solution:-let x be a cube root of unity, then
x^3=1
= x=(1)^1/3 = (1+0i)^1/3
x=(cos0+isin0)^1/3
x=(cos(2nπ)+isin(2nπ))^1/3
x=(cos(2nπ/3)+isin(2nπ/3))
Putting n=0,1,2
x=1, cos2π/3+isin2π/3, cos4π/3+isin4π/3
x=1, cos2π/3+isin2π/3, (cos2π/3+isin2π/3)^2
Putting ⍵ = cos2π/3+isin2π/3
x=1,⍵,⍵^2
since ⍵ is a cube root of unity, then
⍵^3=1
⍵^3-1=0
(⍵-1)(⍵^2+⍵+1)=0
such that ⍵≠1
(⍵^2+⍵+1)=0
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