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

The correct option is A $\frac{\pi }{4}$

We know that $z_0=\omega$ or ${\omega }^2$ (where $\omega$ is a non-real cube root of unity)

$z=3+6i(\omega {)}^{81}-3i(\omega {)}^{93}$

$z=3+3i$ $[\because {\omega }^3=1]$

$z=3+3i$ comparing with $z=x+iy$

$\Rightarrow argz={\tan }^{-1}\left(\frac{y}{x}\right)=\frac{\pi }{4}$