Chapter 8: Problem 96
Solve the equation. $$z^{3}-1=0$$
Short Answer
Expert verified
The solutions are \( z = 1 \), \( z = \frac{-1 + i\sqrt{3}}{2} \), and \( z = \frac{-1 - i\sqrt{3}}{2} \).
Step by step solution
01
Understand the Problem
We need to solve the equation \( z^3 - 1 = 0 \). This equation is asking us to find the values of \( z \) that satisfy this equation.
02
Rewrite the Equation
Recognize \( z^3 - 1 = 0 \) as a difference of cubes. It can be rewritten as \( (z - 1)(z^2 + z + 1) = 0 \).
03
Solve Each Factor Separately
We solve for \( z \) by setting each factor equal to zero: 1. \( z - 1 = 0 \)2. \( z^2 + z + 1 = 0 \).
04
Solve the First Factor
Solve \( z - 1 = 0 \) by adding 1 to both sides, which gives us the solution \( z = 1 \).
05
Solve the Quadratic Part
The solution to \( z^2 + z + 1 = 0 \) can be found using the quadratic formula: \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \).
06
Apply the Quadratic Formula
Substitute into the quadratic formula: \( z = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times 1}}{2 \times 1} = \frac{-1 \pm \sqrt{-3}}{2} \).
07
Simplify the Imaginary Roots
Since the discriminant is negative, we have complex roots: \( z = \frac{-1 \pm i\sqrt{3}}{2} \). This results in two complex solutions: \( z = \frac{-1 + i\sqrt{3}}{2} \) and \( z = \frac{-1 - i\sqrt{3}}{2} \).
08
List All Solutions
Collect all roots: \( z = 1 \), \( z = \frac{-1 + i\sqrt{3}}{2} \), and \( z = \frac{-1 - i\sqrt{3}}{2} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a powerful tool used to find the roots of a quadratic equation, which is a polynomial of the form \( ax^2 + bx + c = 0 \). This formula provides the solution by calculating the values of \( x \) that satisfy the equation. The formula is given by:
A key part of the quadratic formula is the "discriminant" (\( b^2 - 4ac \)). This value reveals a lot about the nature of the roots:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
A key part of the quadratic formula is the "discriminant" (\( b^2 - 4ac \)). This value reveals a lot about the nature of the roots:
- If \( b^2 - 4ac > 0 \), the equation has two distinct real roots.
- If \( b^2 - 4ac = 0 \), it has exactly one real root, also called a repeated or double root.
- If \( b^2 - 4ac < 0 \), the equation has two complex roots.
Complex Roots
Complex roots appear when the discriminant of a quadratic equation is negative. This means the equation cannot be easily solved using just real numbers. In such cases, the roots include imaginary numbers. Imaginary numbers are multiples of \( i \), where \( i \) is the imaginary unit and satisfies \( i^2 = -1 \).
When you use the quadratic formula and encounter a negative discriminant, the square root of the negative discriminant will involve \( i \). For example, with a discriminant of \(-3\), its square root is \( i\sqrt{3} \).
Thus, using this method, the roots become complex numbers and can be expressed in the form:
When you use the quadratic formula and encounter a negative discriminant, the square root of the negative discriminant will involve \( i \). For example, with a discriminant of \(-3\), its square root is \( i\sqrt{3} \).
Thus, using this method, the roots become complex numbers and can be expressed in the form:
- \( z = x + yi \)
Difference of Cubes
The formula for the difference of cubes is a useful identity in algebra, especially when factoring specific polynomial expressions. When you have a difference of cubes, it's in the form \( a^3 - b^3 \). This can be factored using:
In the given exercise, \( z^3 - 1 \) fits the difference of cubes form \( a^3 - b^3 \), with \( a = z \) and \( b = 1 \). By this formula, it factors into:
- \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)
In the given exercise, \( z^3 - 1 \) fits the difference of cubes form \( a^3 - b^3 \), with \( a = z \) and \( b = 1 \). By this formula, it factors into:
- \( (z - 1)(z^2 + z + 1) = 0 \)