Question

Let $z_0$ be a root of the quadratic equation, $x^2+x+1=0$. If $z=3+6i{z}_0^{81}-3i{z}_0^{93}$ , then arg $z$ is equal to :

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

Answer 1

Answer: (A)

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