Solution.
z 3 − α 3 = 0 ; z^3-\alpha^3=0; z 3 − α 3 = 0 ;
z 3 − α 3 = ( z − α ) ( z 2 + α z + α 2 ) ; z^3-\alpha^3=(z-\alpha)(z^2+\alpha z+\alpha^2); z 3 − α 3 = ( z − α ) ( z 2 + α z + α 2 ) ;
( z − α ) ( z 2 + α z + α 2 ) = 0 ; (z-\alpha)(z^2+\alpha z+\alpha^2)=0; ( z − α ) ( z 2 + α z + α 2 ) = 0 ;
z − α = 0 z-\alpha=0 z − α = 0 z 2 + α z + α 2 = 0 ; z^2+\alpha z+\alpha^2=0; z 2 + α z + α 2 = 0 ;
z 1 = α ; z_1=\alpha; z 1 = α ; D = α 2 − 4 α 2 = 3 α 2 ; D=\alpha^2-4\alpha^2=3\alpha^2; D = α 2 − 4 α 2 = 3 α 2 ;
z 2 = − α + 3 α 2 2 = α − 1 + i 3 2 ; z_2=\dfrac{-\alpha+\sqrt{3\alpha^2}}{2}=\alpha \dfrac{-1+i\sqrt{3}}{2}; z 2 = 2 − α + 3 α 2 = α 2 − 1 + i 3 ;
z 3 = − α − 3 α 2 2 = α − 1 − i 3 2 ; z_3=\dfrac{-\alpha-\sqrt{3\alpha^2}}{2}=\alpha \dfrac{-1-i\sqrt{3}}{2}; z 3 = 2 − α − 3 α 2 = α 2 − 1 − i 3 ;
Cube Root of Unity Value:
ω 1 = 1 \omega_1=1 ω 1 = 1
ω 2 = − 1 + i 3 2 \omega_2= \dfrac{-1+i\sqrt{3}}{2} ω 2 = 2 − 1 + i 3
ω 3 = − 1 − i 3 2 \omega_3= \dfrac{-1-i\sqrt{3}}{2} ω 3 = 2 − 1 − i 3
z 1 = α ⋅ ω 1 = α ⋅ 1 = α ; z_1=\alpha\sdot\omega_1=\alpha\sdot1=\alpha; z 1 = α ⋅ ω 1 = α ⋅ 1 = α ;
z 2 = α ⋅ ω 2 = α − 1 + i 3 2 = − 1 2 α + i α 3 2 ; z_2=\alpha\sdot\omega_2=\alpha \dfrac{-1+i\sqrt{3}}{2}=-\dfrac{1}{2}\alpha+i\dfrac{\alpha\sqrt{3}}{2}; z 2 = α ⋅ ω 2 = α 2 − 1 + i 3 = − 2 1 α + i 2 α 3 ;
z 3 = α ⋅ ω 3 = α − 1 − i 3 2 = − 1 2 α − i α 3 2 ; z_3=\alpha\sdot\omega_3=\alpha \dfrac{-1-i\sqrt{3}}{2}=-\dfrac{1}{2}\alpha-i\dfrac{\alpha\sqrt{3}}{2}; z 3 = α ⋅ ω 3 = α 2 − 1 − i 3 = − 2 1 α − i 2 α 3 ;
Answer: z 1 = α ; z_1=\alpha; z 1 = α ;
z 2 = − 1 2 α + i α 3 2 ; z_2=-\dfrac{1}{2}\alpha+i\dfrac{\alpha\sqrt{3}}{2}; z 2 = − 2 1 α + i 2 α 3 ;
z 3 = − 1 2 α − i α 3 2 ; z_3=-\dfrac{1}{2}\alpha-i\dfrac{\alpha\sqrt{3}}{2}; z 3 = − 2 1 α − i 2 α 3 ;
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