MHT CET · Maths · Complex Number
If \(\alpha\) and \(\beta\) are imaginary cube roots of units then value of \(\alpha^4+\beta^{28}+\frac{1}{\alpha \beta}\) is
- A \(1\)
- B \(0\)
- C \(-1\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(0\)
Step-by-step Solution
Detailed explanation
Let \(\alpha=\omega, \beta=\omega^2\)
Now \(\alpha^4+\beta^{28}+\frac{1}{\alpha \beta}\)
\(\begin{aligned} & =\omega^4+\left(\omega^2\right)^{28}+\frac{1}{\omega \omega^2} \\ & =\omega^4+\omega^{56}+\frac{1}{\omega^3} \\ & =\omega+\omega^2+\frac{1}{1} \\ & \left.=0 \quad \quad \quad \text { [as } \omega^3=1\right]\end{aligned}\)
Now \(\alpha^4+\beta^{28}+\frac{1}{\alpha \beta}\)
\(\begin{aligned} & =\omega^4+\left(\omega^2\right)^{28}+\frac{1}{\omega \omega^2} \\ & =\omega^4+\omega^{56}+\frac{1}{\omega^3} \\ & =\omega+\omega^2+\frac{1}{1} \\ & \left.=0 \quad \quad \quad \text { [as } \omega^3=1\right]\end{aligned}\)
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