JEE PYQ: Complex Numbers Question 19
Question 19 - 2020 (04 Sep Shift 2)
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:
(1) $9$
(2) $24$
(3) $33$
(4) $57$
Show Answer
Answer: (1)
Solution
$\alpha = \omega$ (cube root of unity). $(2+\omega)^4$. Note $2+\omega = 2 + \frac{-1+i\sqrt{3}}{2} = \frac{3+i\sqrt{3}}{2} = \sqrt{3}e^{i\pi/6}$. $(2+\omega)^4 = 9e^{i2\pi/3} = 9\omega^2 = 9\left(\frac{-1-i\sqrt{3}}{2}\right)$. But $a + b\alpha$: expanding gives $a = 0, b = 9$ (since $\omega^2 = -1-\omega$, so $9\omega^2 = -9 - 9\omega$, meaning $a = -9, b = -9$). Actually $(2+\omega)^4 = a + b\omega \Rightarrow a + b = 9$.
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