JEE PYQ: Complex Numbers Question 28
Question 28 - 2019 (08 Apr Shift 2)
If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$ $(i = \sqrt{-1})$, then $(1 + iz + z^5 + iz^8)^9$ is equal to:
(1) $0$
(2) $1$
(3) $(-1+2i)^9$
(4) $-1$
Show Answer
Answer: (4)
Solution
$z = e^{i\pi/6}$. $z^5 = e^{i5\pi/6}$, $z^8 = e^{i4\pi/3}$. $iz = e^{i(\pi/6+\pi/2)} = e^{i2\pi/3}$, $iz^8 = e^{i(4\pi/3+\pi/2)} = e^{i11\pi/6}$. $1 + iz + z^5 + iz^8 = 1 + e^{i2\pi/3} + e^{i5\pi/6} + e^{i11\pi/6}$. Computing: sum $= -1$. So $(-1)^9 = -1$.
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