If A = {x ϵ Z : x 3– 1 = 0} and B = {x ϵ Z: x 2+ x + 1 = 0}, where Z is set of complex numbers, then what is A ∩ B equal to?
Understanding the Problem: Set Intersection and Complex Roots
The question asks us to find the intersection of two sets, A and B. Both sets are defined by equations whose solutions are complex numbers. Set A contains the roots of the equation \(x^3 - 1 = 0\), and set B contains the roots of the equation \(x^2 + x + 1 = 0\). The elements of both sets are specified to be within the set Z, which the question defines as the set of complex numbers.
To find the intersection \(A \cap B\), we first need to determine the elements (roots) present in each set.
Finding the Elements of Set A: Roots of \(x^3 - 1 = 0\)
Set A is defined as \(A = \{x \in \mathbb{Z} : x^3 - 1 = 0\}\). We need to find the roots of the equation \(x^3 - 1 = 0\). This is a standard equation for finding the cube roots of unity.
We can factor the equation \(x^3 - 1 = 0\) using the difference of cubes formula, \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\):
\[(x-1)(x^2 + x + 1) = 0\]
This gives us two possibilities for the roots:
- \(x-1 = 0 \implies x = 1\)
- \(x^2 + x + 1 = 0\)
The first root is \(x=1\). To find the roots from the second equation, \(x^2 + x + 1 = 0\), we use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a=1\), \(b=1\), and \(c=1\).
\[x = \frac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2(1)}\]
\[x = \frac{-1 \pm \sqrt{1 - 4}}{2}\]
\[x = \frac{-1 \pm \sqrt{-3}}{2}\]
\[x = \frac{-1 \pm i\sqrt{3}}{2}\]
So the other two roots are \(x = \frac{-1 + i\sqrt{3}}{2}\) and \(x = \frac{-1 - i\sqrt{3}}{2}\).
Since Z is the set of complex numbers, all these roots belong to Z. Therefore, the elements of set A are:
\[A = \left\{1, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}\right\}\]
Finding the Elements of Set B: Roots of \(x^2 + x + 1 = 0\)
Set B is defined as \(B = \{x \in \mathbb{Z} : x^2 + x + 1 = 0\}\). We need to find the roots of the equation \(x^2 + x + 1 = 0\).
As we already solved this equation when finding the roots of \(x^3 - 1 = 0\), we know the roots are:
\[x = \frac{-1 + i\sqrt{3}}{2} \quad \text{and} \quad x = \frac{-1 - i\sqrt{3}}{2}\]
Since Z is the set of complex numbers, these roots belong to Z. Therefore, the elements of set B are:
\[B = \left\{\frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}\right\}\]
Determining the Intersection \(A \cap B\)
The intersection of two sets, \(A \cap B\), is the set of elements that are common to both A and B.
We have:
- \(A = \left\{1, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}\right\}\)
- \(B = \left\{\frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}\right\}\)
Comparing the elements of A and B, we can see which elements are present in both sets. The elements \(\frac{-1 + i\sqrt{3}}{2}\) and \(\frac{-1 - i\sqrt{3}}{2}\) are present in both set A and set B.
The element \(1\) is in set A but not in set B.
Therefore, the intersection \(A \cap B\) is the set containing the common elements:
\[A \cap B = \left\{\frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}\right\}\]
This result matches one of the given options.
| Set | Defining Equation | Elements (Complex Roots) |
|---|---|---|
| A | \(x^3 - 1 = 0\) | \(1, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}\) |
| B | \(x^2 + x + 1 = 0\) | \(\frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}\) |
The elements common to A and B form the intersection \(A \cap B\).
Conclusion on Set Intersection
Based on our analysis, the intersection of set A and set B is the set containing the complex roots of the equation \(x^2 + x + 1 = 0\).
\[A \cap B = \left\{\frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}\right\}\]
Revision Table: Set Theory and Complex Numbers
| Concept | Description | Example |
|---|---|---|
| Set Intersection (\(A \cap B\)) | The set containing elements that are in both set A and set B. | If \(A = \{1, 2, 3\}\) and \(B = \{2, 3, 4\}\), then \(A \cap B = \{2, 3\}\). |
| Complex Number | A number of the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit (\(i^2 = -1\)). | \(3 + 4i\), \(-1 - \sqrt{3}i\) |
| Roots of Unity | Complex numbers that satisfy the equation \(x^n = 1\) for a positive integer \(n\). | Cube roots of unity are the roots of \(x^3 = 1\). |
| Quadratic Formula | Formula to find the roots of a quadratic equation \(ax^2 + bx + c = 0\): \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). | Used to solve \(x^2 + x + 1 = 0\). |
Additional Information: Properties of Complex Roots
The complex roots \(\frac{-1 + i\sqrt{3}}{2}\) and \(\frac{-1 - i\sqrt{3}}{2}\) are often denoted by \(\omega\) and \(\omega^2\) (or vice versa), and they are the non-real cube roots of unity. They have interesting properties:
- \(\omega^2 + \omega + 1 = 0\)
- \(\omega^3 = 1\)
- The roots of \(x^3 - 1 = 0\) are \(1, \omega, \omega^2\).
- The roots of \(x^2 + x + 1 = 0\) are \(\omega, \omega^2\).
Understanding these properties can simplify problems involving cube roots of unity.
In this question, set A consists of the cube roots of unity, and set B consists of the non-real cube roots of unity. The intersection \(A \cap B\) is simply the set of elements common to both, which are the non-real cube roots of unity.
Important Questions from Union and Intersection of Sets
Consider the following statements
1. A = (A ∪ B) ∪ (A - B),
2. A ∪ (B - A) = (A ∪ B)
3. B = (A ∪ B) - (A - B)
Which of the statements given above are correct?
If a set A contains 3 elements and another set B contains 6 elements, then what is the minimum number of elements that (A∪B) can have?
If A = {x : 0 ≤ x ≤ 2} and B = {y; y is a prime number}, then what is A ∩ B equal to?
Let A ∪ B = {x|(x - a)(x - b) > 0, where a < b}. What are A and B equal to?
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper I, 26 read newspaper T, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find the number of students who read exactly one newspaper.
