Let z_0 be a root of x^2+x+1=0.

Question

Let $z_0$ be a root of $x^2+x+1=0$. If $z=3+6iz_0^{81}-3iz_0^{93}$, then $\arg z$ is equal to:

Accepted Answer

$z_0=\omega$ or $\omega^2$, so $z_0^3=1$. Then $z=3+6i(z_0^3)^{27}-3i(z_0^3)^{31}=3+6i-3i=3+3i$.

$$\arg z=\tan^{-1}\frac33=\frac\pi4.$$

Solutions · 1