1. Roots of the quadratic

The equation is $x^2+x+1=0.$ Its roots are the non-real cube roots of unity: $z_0=\omega \quad \text{or} \quad \omega^2,$ where $\omega^3=1, \qquad \omega\ne 1, \qquad 1+\omega+\omega^2=0.$ So for any root $z_0$ of this equation, $z_0^3=1.$

2. Reduce the powers $81$ and $93$

Given $z=3+6iz_0^{81}-3iz_0^{93}.$ Since $z_0^3=1$, powers repeat modulo $3$.

Now, $81\equiv 0 \pmod 3, \qquad 93\equiv 0 \pmod 3.$ Hence, $z_0^{81}=(z_0^3)^{27}=1,$ $z_0^{93}=(z_0^3)^{31}=1.$

Therefore, $z=3+6i(1)-3i(1)=3+3i.$

3. Find the argument

For $z=3+3i,$ we have real part $=3$ and imaginary part $=3$, so $z$ lies in the first quadrant.

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

4. Check options

• A: $\frac{\pi}{4}$ ✅
• B: $\frac{\pi}{6}$ ❌
• C: $\frac{\pi}{3}$ ❌
• D: $0$ ❌

So the correct answer is A.