(C) Given,$\alpha^{2}-\alpha+1=0$. Multiplying by $(\alpha+1)$,we get $(\alpha+1)(\alpha^{2}-\alpha+1)=0$,which implies $\alpha^{3}+1=0$,so $\alpha^{3

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(C) Given,$\alpha^{2}-\alpha+1=0$. Multiplying by $(\alpha+1)$,we get $(\alpha+1)(\alpha^{2}-\alpha+1)=0$,which implies $\alpha^{3}+1=0$,so $\alpha^{3}=-1$. Therefore,$\alpha^{6}=1$. We need to find $\alpha^{2011}$. Since $2011 = 6 \times 335 + 1$,we have $\alpha^{2011} = (\alpha^{6})^{335} \times \alpha^{1} = (1)^{335} \times \alpha = \alpha$. Thus,$\alpha^{2011} = \alpha$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers De Moivre's theorem and Roots of unity

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If $\alpha$ is a complex number such that $\alpha^{2}-\alpha+1=0$,then $\alpha^{2011}$ is equal to

KCET 2012 All KCET

A
$-\alpha$
B
$\alpha^{2}$
C
$\alpha$
D
$1$

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1 Complex Numbers and Quadratic Equations: Complex Numbers

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De Moivre's theorem and Roots of unity

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