Let z_0 be a root of the quadratic equation,x | vedclass

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(A) The roots of the quadratic equation $x^2 + x + 1 = 0$ are the complex cube roots of unity,$\omega$ and $\omega^2$.Given $z = 3 + 6iz_0^{81} - 3iz_

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$\frac{\pi}{4}$

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(A) The roots of the quadratic equation $x^2 + x + 1 = 0$ are the complex cube roots of unity,$\omega$ and $\omega^2$.Given $z = 3 + 6iz_0^{81} - 3iz_0^{93}$.Since $\omega^3 = 1$,we have $z_0^{81} = (z_0^3)^{27} = 1^{27} = 1$ and $z_0^{93} = (z_0^3)^{31} = 1^{31} = 1$.Substituting these values into the expression for $z$:$z = 3 + 6i(1) - 3i(1) = 3 + 3i$.To find $\arg(z)$,we use the formula $\arg(x + iy) = 	an^{-1}(\frac{y}{x})$ for $x > 0$.$\arg(z) = 	an^{-1}(\frac{3}{3}) = 	an^{-1}(1) = \frac{\pi}{4}$.

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

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