Let Z0 be a root of the quadratic equation,


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

  1. (1) $\frac{\pi}{4}$

  2. (2) $\frac{\pi}{6}$

  3. (3) $\frac{\pi}{3}$

  4. (4) 0

Correct Option: 1


$\because z_{0}$ is a root of quadratic equation


$\therefore \quad z_{0}=\omega$ or $\omega^{2} \Rightarrow z_{0}^{3}=1$

$\therefore \quad z=3+6 i z_{0}^{81}-3 i z_{0}^{93}$

$=3+6 i\left(\left(z_{0}\right)^{3}\right)^{27}-3 i\left(\left(z_{0}\right)^{3}\right)^{31}$

$=3+6 i-3 i$

$=3+3 i$

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


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