Question

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 :

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{3}$
• C. 0
• D. $\frac{\pi}{6}$

Answer 1

Answer: (A)

$\frac{\pi}{4}$

Answer 2

$z_0 = \omega$ or $\omega^2$ (where $\omega$ is a non-real cube root of unity)
$z = 3 + 6i(\omega)^{81} - 3i(\omega)^{93}$
z = 3 + 3i
$\Rightarrow \; \arg \; z = \frac{\pi}{4}$