A) \[\alpha \]
B) \[{{\alpha }^{2}}\]
C) \[1\]
D) \[i\]
Correct Answer: A
Solution :
Given equation is \[{{\alpha }^{2}}+\alpha +1=0\] \[\therefore \]\[\alpha =\frac{-1\pm \sqrt{1-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}\] Let it be \[\alpha =\omega ,\,{{\omega }^{2}}\] (1) If \[\alpha =\omega ,\,\] then \[{{\alpha }^{31}}={{(\omega )}^{31}}=\omega =\alpha \] (2) If \[\alpha ={{\omega }^{2}},\] then \[{{\alpha }^{31}}={{({{\omega }^{2}})}^{31}}={{\omega }^{62}}={{\omega }^{2}}=\alpha \] Hence, \[{{\alpha }^{31}}\] is equal to a.
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