Animated Solution for Mathematics - Complex Numbers: 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:

When you encounter the quadratic equation $x^2+x+1=0$, pause and recognize it as the signature of the cube roots of unity. This equation serves as a gateway to a world where numbers rotate and dance on the complex plane.

The roots of this equation are the non-real cube roots of unity, denoted as $\omega$ and ${\omega }^2$. The most fundamental property of these roots is that ${\omega }^3=1$.

This property acts as our master key. Whenever you encounter a high power of $\omega$, you are essentially looking at a cycle of three; if the exponent is a multiple of 3, the result is simply 1.

Consider the expression:

We must evaluate $z_0^{81}$ and $z_0^{93}$. Since $81=3\times 27$, we have:

Similarly, since $93=3\times 31$, we find:

The complex powers vanish, leaving us with a simplified expression.

Substitute these values back into the equation for $z$:

This simplifies to:

Imagine this point on the complex plane. You move 3 units along the real axis and 3 units up the imaginary axis, placing you in the first quadrant at coordinates $(3,3)$. This represents a vector pointing exactly at a 45-degree angle.

To find the argument, we use the standard formula for a complex number $z=x+iy$ located in the first quadrant:

Given $x=3$ and $y=3$, we calculate:

We know that the angle whose tangent is 1 is $\frac{\pi }{4}$. Thus, the final argument is:

We started with a complex-looking expression and, by understanding the underlying structure of the roots of unity, we arrived at a simple, elegant result. Keep this perspective, and no complex number problem will ever intimidate you again.