JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(\alpha\) be a root of the equation \(1+x^{2}+x^{4}=0\). Then the value of \(\alpha^{1011}+\alpha^{\text {2022 }}-\alpha^{\text {3033}}\) is equal to
- A \(1\)
- B \(\alpha\)
- C \(1+\alpha\)
- D \(1+2 \alpha\)
Answer & Solution
Correct Answer
(A) \(1\)
Step-by-step Solution
Detailed explanation
\(x^{4}+x^{2}+1=0\) \(\Rightarrow\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)=0\) \(\Rightarrow x=\pm \;\omega, \pm \;\omega^{2}\) where \(\omega=1^{1 / 3}\) and imaginary. So \(\alpha^{1011}+\alpha^{2002}-\alpha^{3033}=1+1-1=1\)
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