JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(\alpha ,\beta \in C\) are distinct roots, of the equatin \({x^2} - x + 1 = 0\) ,then \({\alpha ^{101}} + {\beta ^{107}}\) is equal to :
- A \(0\)
- B \(1\)
- C \(2\)
- D \(-1\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(\alpha, \beta\) are roots of \(x^{2}-x+1=0\) \(\therefore \quad \alpha=-\omega\) and \(\beta=-\omega^{2}\) where \(\omega\) is cube root of unity \(\therefore \quad \alpha^{101}+\beta^{107}\) \(=(-\omega)^{101}+(-\omega)^{107}\) \(=-\left[\omega^{2}+\omega\right]=-[-1]=1\)
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