If `z^(2)+z+1=0` then the value of
`(z+1/z)^(2)+(z^(2)+1/z^(2))^(2)+(z^(3)+1/z^(3))^(2)+....+(z^(21)+1/z^(21))^(2)` is equal to

Options

A. 21
B. 42
C. 0
D. 11

Answer

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

\[ (z + \frac{1}{z})^2 + (z^2 + \frac{1}{z^2})^2 + (z^3 + \frac{1}{z^3})^2 + \ldots + (z^{21} + \frac{1}{z^{21}})^2, \]

we will follow these steps:

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

Using the quadratic formula:

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

Thus, the roots are:

\[ z_1 = \frac{-1 + i\sqrt{3}}{2}, \quad z_2 = \frac{-1 - i\sqrt{3}}{2}. \]

These roots can be recognized as the cube roots of unity, specifically \( \omega \) and \( \omega^2 \), where \( \omega = e^{2\pi i / 3} \).

### Step 2: 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 3: Calculate \( z^2 + \frac{1}{z^2} \)

Using the identity:

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

### Step 4: Calculate \( z^3 + \frac{1}{z^3} \)

Using the identity:

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

### Step 5: Generalize for \( z^n + \frac{1}{z^n} \)

For \( n \geq 1 \):

- If \( n \equiv 0 \mod 3 \): \( z^n + \frac{1}{z^n} = 2 \)
- If \( n \equiv 1 \mod 3 \): \( z^n + \frac{1}{z^n} = -1 \)
- If \( n \equiv 2 \mod 3 \): \( z^n + \frac{1}{z^n} = -1 \)

### Step 6: Calculate the sum

The series \( (z^n + \frac{1}{z^n})^2 \):

- For \( n \equiv 0 \mod 3 \): \( (2)^2 = 4 \)
- For \( n \equiv 1 \mod 3 \): \( (-1)^2 = 1 \)
- For \( n \equiv 2 \mod 3 \): \( (-1)^2 = 1 \)

### Step 7: Count the terms from \( n = 1 \) to \( n = 21 \)

The sequence from 1 to 21 has:

- 7 terms where \( n \equiv 0 \mod 3 \) (3, 6, 9, 12, 15, 18, 21)
- 7 terms where \( n \equiv 1 \mod 3 \) (1, 4, 7, 10, 13, 16, 19)
- 7 terms where \( n \equiv 2 \mod 3 \) (2, 5, 8, 11, 14, 17, 20)

### Step 8: Calculate the total sum

\[ S = 7 \cdot 4 + 7 \cdot 1 + 7 \cdot 1 = 28 + 7 + 7 = 42. \]

### Final Answer

The value of

\[ (z + \frac{1}{z})^2 + (z^2 + \frac{1}{z^2})^2 + (z^3 + \frac{1}{z^3})^2 + \ldots + (z^{21} + \frac{1}{z^{21}})^2 = 42. \]