If `A={x inZ:x^(2)-1=0}and B={x inZ:x^(2)+x+1=0},` where Z is set of complex numbers, then what is `A sum B` equal to ?

Options

A. Null set
B. `{(-1+sqrt3i)/(2),(1-sqrt3i)/(2)}`
C. `{(-1+sqrt3i)/(4),(-1-sqrt3i)/(4)}`
D. `{(1+sqrt3i)/(2),(1-sqrt3i)/(2)}`

Explanation

To solve the problem, we need to find the sets \( A \) and \( B \) based on the given equations, and then determine the sum of these sets, denoted as \( A \sum B \).

### Step 1: Define Set \( A \)
Set \( A \) is defined as:
\[ A = \{ x \in \mathbb{Z} : x^2 - 1 = 0 \} \]
To find the elements of set \( A \), we solve the equation:
\[ x^2 - 1 = 0 \]
Factoring gives:
\[ (x - 1)(x + 1) = 0 \]
Thus, the solutions are:
\[ x = 1 \quad \text{or} \quad x = -1 \]
So, we have:
\[ A = \{ 1, -1 \} \]

### Step 2: Define Set \( B \)
Set \( B \) is defined as:
\[ B = \{ x \in \mathbb{Z} : x^2 + x + 1 = 0 \} \]
To find the elements of set \( B \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = 1 \). Plugging in these values gives:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} \]
This simplifies to:
\[ x = \frac{-1 \pm i\sqrt{3}}{2} \]
Thus, the roots are:
\[ x = \frac{-1 + i\sqrt{3}}{2} \quad \text{and} \quad x = \frac{-1 - i\sqrt{3}}{2} \]
So, we have:
\[ B = \left\{ \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2} \right\} \]

### Step 3: Calculate \( A \sum B \)
The sum of the sets \( A \) and \( B \) is defined as:
\[ A \sum B = \{ a + b : a \in A, b \in B \} \]
Calculating each combination:
1. For \( a = 1 \):
- \( 1 + \frac{-1 + i\sqrt{3}}{2} = \frac{2}{2} + \frac{-1 + i\sqrt{3}}{2} = \frac{1 + i\sqrt{3}}{2} \)
- \( 1 + \frac{-1 - i\sqrt{3}}{2} = \frac{2}{2} + \frac{-1 - i\sqrt{3}}{2} = \frac{1 - i\sqrt{3}}{2} \)

2. For \( a = -1 \):
- \( -1 + \frac{-1 + i\sqrt{3}}{2} = \frac{-2}{2} + \frac{-1 + i\sqrt{3}}{2} = \frac{-3 + i\sqrt{3}}{2} \)
- \( -1 + \frac{-1 - i\sqrt{3}}{2} = \frac{-2}{2} + \frac{-1 - i\sqrt{3}}{2} = \frac{-3 - i\sqrt{3}}{2} \)

Thus, the final set \( A \sum B \) is:
\[ A \sum B = \left\{ \frac{1 + i\sqrt{3}}{2}, \frac{1 - i\sqrt{3}}{2}, \frac{-3 + i\sqrt{3}}{2}, \frac{-3 - i\sqrt{3}}{2} \right\} \]

### Final Answer
\[ A \sum B = \left\{ \frac{1 + i\sqrt{3}}{2}, \frac{1 - i\sqrt{3}}{2}, \frac{-3 + i\sqrt{3}}{2}, \frac{-3 - i\sqrt{3}}{2} \right\} \]