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` is

Options

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

Answer

To solve the equation \( z^2 + z + 1 = 0 \) and find the value of \( (z + \frac{1}{z})^2 + (z^2 + \frac{1}{z^2})^2 + \ldots + (z^6 + \frac{1}{z^6})^2 \), we can follow these steps:

### Step 1: Solve the quadratic equation
We start with the quadratic equation:
\[ z^2 + z + 1 = 0 \]
Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, c = 1 \):
\[ z = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} \]
This simplifies to:
\[ z = \frac{-1 \pm i\sqrt{3}}{2} \]
These roots can be represented as \( \omega \) and \( \omega^2 \), where \( \omega = e^{i\frac{\pi}{3}} \) and \( \omega^2 = e^{-i\frac{\pi}{3}} \).

### Step 2: Calculate \( z + \frac{1}{z} \)
Using the property of \( z \):
\[ z + \frac{1}{z} = z + \frac{1}{\frac{-1 \pm i\sqrt{3}}{2}} = z + \frac{2}{-1 \pm i\sqrt{3}} \]
Multiplying numerator and denominator by the conjugate:
\[ z + \frac{1}{z} = z + \frac{2(-1 \mp i\sqrt{3})}{(-1)^2 + (i\sqrt{3})^2} = z + \frac{2(-1 \mp i\sqrt{3})}{1 + 3} = z + \frac{-1 \mp i\sqrt{3}}{2} \]
This simplifies to:
\[ z + \frac{1}{z} = -1 \]

### Step 3: Calculate \( (z + \frac{1}{z})^2 \)
Now we calculate:
\[ (z + \frac{1}{z})^2 = (-1)^2 = 1 \]

### Step 4: Calculate \( z^2 + \frac{1}{z^2} \)
Using the identity:
\[ z^2 + \frac{1}{z^2} = (z + \frac{1}{z})^2 - 2 = 1 - 2 = -1 \]
Then,
\[ (z^2 + \frac{1}{z^2})^2 = (-1)^2 = 1 \]

### Step 5: Calculate \( z^3 + \frac{1}{z^3} \)
Using the identity:
\[ z^3 + \frac{1}{z^3} = (z + \frac{1}{z})(z^2 + \frac{1}{z^2}) - (z + \frac{1}{z}) = (-1)(-1) - (-1) = 1 + 1 = 2 \]
Then,
\[ (z^3 + \frac{1}{z^3})^2 = 2^2 = 4 \]

### Step 6: Continue for \( z^4, z^5, z^6 \)
Following the same pattern:
- \( z^4 + \frac{1}{z^4} = -1 \) leads to \( (z^4 + \frac{1}{z^4})^2 = 1 \)
- \( z^5 + \frac{1}{z^5} = 2 \) leads to \( (z^5 + \frac{1}{z^5})^2 = 4 \)
- \( z^6 + \frac{1}{z^6} = 1 \) leads to \( (z^6 + \frac{1}{z^6})^2 = 1 \)

### Step 7: Sum all the squares
Now we sum up all the squares:
\[ 1 + 1 + 4 + 1 + 4 + 1 = 12 \]

### Final Answer
Thus, the value of \( (z + \frac{1}{z})^2 + (z^2 + \frac{1}{z^2})^2 + \ldots + (z^6 + \frac{1}{z^6})^2 \) is:
\[ \boxed{12} \]