Question

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 :

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{3}$
• C. 0
• D. $\frac{\pi}{6}$

Answer 1

The Correct Option is A

To solve the given problem, we start by analyzing the quadratic equation \(x^2 + x + 1 = 0\). The roots of this equation are given by the formula for the roots of a quadratic equation:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

For the equation \(x^2 + x + 1 = 0\), we have \(a = 1\), \(b = 1\), and \(c = 1\). Let's calculate the discriminant:

\(\Delta = b^2 - 4ac = 1^2 - 4 \times 1 \times 1 = 1 - 4 = -3.\)

Since the discriminant is negative, the roots are complex, given by:

\(x = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}\)

Let's denote the complex roots as \(z_0 = \omega = \frac{-1 + i\sqrt{3}}{2}\) and \(\omega^2 = \frac{-1 - i\sqrt{3}}{2}\). These are the cube roots of unity, other than 1 itself, i.e., \(\omega^3 = 1\).

We know that:

\(\omega^3 = 1 \Rightarrow \omega^n = \omega^{n \mod 3}.\)

Now, let's calculate \(z_0^{81}\) and \(z_0^{93}\):

1. \(z_0^{81} = (\omega)^{81} = (\omega^3)^{27} = 1^{27} = 1\).
2. \(z_0^{93} = (\omega)^{93} = (\omega^3)^{31} = 1^{31} = 1\).

Substitute these into the expression for \(z\):

\(z = 3 + 6i \cdot 1 - 3i \cdot 1 = 3 + 6i - 3i = 3 + 3i.\)

Now, let's find the argument of \(z\):

\(= 3 + 3i\) is of form \(a + bi\) with \(a = 3, b = 3\). The argument \(\text{arg}(z)\) i\)

\(\text{arg}(z) = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{3}{3}\right) = \tan^{-1}(1) = \frac{\pi}{4}.\)

Hence, the argument of \(z\) is \(\boxed{\frac{\pi}{4}}\).