Question:
If a and b are real numbers such that
\((2+\alpha)^4 = a + b\alpha\)
where \(\alpha = \frac{-1+i\sqrt{3}}{2}\)
then a + b is equal to :
If a and b are real numbers such that
\((2+\alpha)^4 = a + b\alpha\)
where \(\alpha = \frac{-1+i\sqrt{3}}{2}\)
then a + b is equal to :
\((2+\alpha)^4 = a + b\alpha\)
where \(\alpha = \frac{-1+i\sqrt{3}}{2}\)
then a + b is equal to :
Show Hint
\(\alpha\) is a primitive cube root of unity, so \(1+\alpha+\alpha^2=0\) and \(\alpha^3=1\). Square \(2+\alpha\) twice.
Updated On: Nov 5, 2026
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The Correct Option is B
Solution and Explanation
\(\alpha = \omega\) is a complex cube root of unity: \(\omega^3 = 1\), \(1+\omega+\omega^2 = 0\).
\((2+\omega)^2 = 4 + 4\omega + \omega^2 = 4+4\omega+(-1-\omega) = 3+3\omega = 3(1+\omega) = -3\omega^2\).
\((2+\omega)^4 = 9\omega^4 = 9\omega\).
Comparing with \(a + b\omega\) (a, b real, and \(1,\omega\) are linearly independent over the reals): \(a = 0,\ b = 9\).
\(a+b = 9\).
Answer: (2) 9
\((2+\omega)^2 = 4 + 4\omega + \omega^2 = 4+4\omega+(-1-\omega) = 3+3\omega = 3(1+\omega) = -3\omega^2\).
\((2+\omega)^4 = 9\omega^4 = 9\omega\).
Comparing with \(a + b\omega\) (a, b real, and \(1,\omega\) are linearly independent over the reals): \(a = 0,\ b = 9\).
\(a+b = 9\).
Answer: (2) 9
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