If ` z^(2) + z + 1 ` = 0 where z is a complex number then the value of ` ( z + (1)/( z))^(2) + ( z^(2) + (1)/( z^(2)))^(2) + . . . + ( z^(6) + (1)/( z^(6))) ^(2) = `

Options

A. 6
B. 12
C. 18
D. 54

Answer

To solve the equation \( z^2 + z + 1 = 0 \) and find the value of the expression

\[ (z + \frac{1}{z})^2 + (z^2 + \frac{1}{z^2})^2 + (z^3 + \frac{1}{z^3})^2 + (z^4 + \frac{1}{z^4})^2 + (z^5 + \frac{1}{z^5})^2 + (z^6 + \frac{1}{z^6})^2, \]

we will follow these steps:

### Step 1: Solve the quadratic equation
The roots of the equation \( z^2 + z + 1 = 0 \) can be found using the quadratic formula:

\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm \sqrt{3}i}{2}. \]

Let \( z_1 = \frac{-1 + \sqrt{3}i}{2} \) and \( z_2 = \frac{-1 - \sqrt{3}i}{2} \).

### Step 2: Identify the roots as cube roots of unity
The roots \( z_1 \) and \( z_2 \) can be expressed as:

\[ z_1 = \omega, \quad z_2 = \omega^2, \]

where \( \omega = e^{2\pi i / 3} \) is a primitive cube root of unity. It satisfies \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

### Step 3: Calculate \( z + \frac{1}{z} \)
For \( z = \omega \):

\[ \frac{1}{z} = \frac{1}{\omega} = \omega^2. \]

Thus,

\[ z + \frac{1}{z} = \omega + \omega^2 = -1. \]

### Step 4: Calculate \( z^n + \frac{1}{z^n} \) for \( n = 1, 2, \ldots, 6 \)
Using the properties of \( \omega \):

- For \( n = 1 \):
\[ z + \frac{1}{z} = -1. \]

- For \( n = 2 \):
\[ z^2 + \frac{1}{z^2} = (\omega^2 + \omega) = -1. \]

- For \( n = 3 \):
\[ z^3 + \frac{1}{z^3} = 1 + 1 = 2. \]

- For \( n = 4 \):
\[ z^4 + \frac{1}{z^4} = \omega + \omega^2 = -1. \]

- For \( n = 5 \):
\[ z^5 + \frac{1}{z^5} = \omega^2 + \omega = -1. \]

- For \( n = 6 \):
\[ z^6 + \frac{1}{z^6} = 1 + 1 = 2. \]

### Step 5: Calculate the squares of these sums
Now we compute the squares:

\[ (z + \frac{1}{z})^2 = (-1)^2 = 1, \]
\[ (z^2 + \frac{1}{z^2})^2 = (-1)^2 = 1, \]
\[ (z^3 + \frac{1}{z^3})^2 = (2)^2 = 4, \]
\[ (z^4 + \frac{1}{z^4})^2 = (-1)^2 = 1, \]
\[ (z^5 + \frac{1}{z^5})^2 = (-1)^2 = 1, \]
\[ (z^6 + \frac{1}{z^6})^2 = (2)^2 = 4. \]

### Step 6: Sum the squares
Now we sum these values:

\[ 1 + 1 + 4 + 1 + 1 + 4 = 12. \]

### Final Answer
Thus, the value of

\[ (z + \frac{1}{z})^2 + (z^2 + \frac{1}{z^2})^2 + (z^3 + \frac{1}{z^3})^2 + (z^4 + \frac{1}{z^4})^2 + (z^5 + \frac{1}{z^5})^2 + (z^6 + \frac{1}{z^6})^2 = 12. \]