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

Answer 1

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

Roots of $x^2+x+1=0$ are $\omega$ and ${\omega }^2$, where $\omega$ is the cube root of unity.
Let $z_0=\omega$
Thus, $z=3+6i(\omega {)}^{81}-3i(\omega {)}^{93}$
$=3+6i-3i[\because {\omega }^3=1]$
$=3+3i$

$\arg \left(z\right)={\tan }^{-1}\left(\frac{3}{3}\right)={\tan }^{-1}\left(1\right)$
$\Rightarrow \arg \left(z\right)=\frac{\pi }{4}$