JEE PYQ: Complex Numbers Question 37
Question 37 - 2019 (09 Jan Shift 2)
Let $z_0$ be a root of the quadratic equation, $x^2 + x + 1 = 0$. If $z = 3 + 6iz_0^{81} - 3iz_0^{93}$, then $\arg z$ is equal to:
(1) $\frac{\pi}{4}$
(2) $\frac{\pi}{6}$
(3) $\frac{\pi}{3}$
(4) $0$
Show Answer
Answer: (1)
Solution
$z_0 = \omega$ (cube root of unity). $z_0^{81} = \omega^{81} = 1$, $z_0^{93} = \omega^{93} = 1$. $z = 3 + 6i - 3i = 3 + 3i$. $\arg(z) = \frac{\pi}{4}$.
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