If w is a complex cube root of unity ,prove that 1+w +w^2 =0

                                       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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