Let `Z_0` is the root of equation `x^2+x+1=0` and `Z=3+6i(Z_0)^(81)-3i(Z_0)^(93)` Then arg `(Z)` is equal to (a) `(pi)/(4)` (b) `(pi)/(3)` (c) `pi` (d) `(pi)/(6)`

Options

A. `(pi)/(4)`
B. `(pi)/(3)`
C. 0
D. `(pi)/(6)`

Answer

To solve the problem, we need to find the argument of the complex number \( Z \) defined as:

\[ Z = 3 + 6i Z_0^{81} - 3i Z_0^{93} \]

where \( Z_0 \) is a root of the equation \( x^2 + x + 1 = 0 \).

### Step 1: Find the roots of the equation \( x^2 + x + 1 = 0 \)

To find the roots, we can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \):

\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]

Thus, the roots are:

\[ Z_0 = \frac{-1 + i\sqrt{3}}{2} \quad \text{and} \quad Z_0^2 = \frac{-1 - i\sqrt{3}}{2} \]

### Step 2: Express \( Z_0 \) in exponential form

The root \( Z_0 \) can be expressed in polar form. The modulus \( r \) of \( Z_0 \) is:

\[ r = \sqrt{\left(\frac{-1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]

The argument \( \theta \) of \( Z_0 \) is:

\[ \theta = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}}\right) = \tan^{-1}(-\sqrt{3}) = \frac{2\pi}{3} \quad (\text{since it is in the second quadrant}) \]

Thus, we can write:

\[ Z_0 = e^{i\frac{2\pi}{3}} \]

### Step 3: Calculate \( Z_0^{81} \) and \( Z_0^{93} \)

Using the property of exponents:

\[ Z_0^{81} = \left(e^{i\frac{2\pi}{3}}\right)^{81} = e^{i\frac{2\pi \cdot 81}{3}} = e^{i54\pi} = e^{i0} = 1 \]

\[ Z_0^{93} = \left(e^{i\frac{2\pi}{3}}\right)^{93} = e^{i\frac{2\pi \cdot 93}{3}} = e^{i62\pi} = e^{i0} = 1 \]

### Step 4: Substitute back into \( Z \)

Now substituting back into the expression for \( Z \):

\[ Z = 3 + 6i(1) - 3i(1) = 3 + 6i - 3i = 3 + 3i \]

### Step 5: Find the argument of \( Z \)

The argument of a complex number \( Z = x + yi \) is given by:

\[ \text{arg}(Z) = \tan^{-1}\left(\frac{y}{x}\right) \]

For \( Z = 3 + 3i \):

\[ \text{arg}(Z) = \tan^{-1}\left(\frac{3}{3}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]

### Final Answer

Thus, the argument of \( Z \) is:

\[ \text{arg}(Z) = \frac{\pi}{4} \]