Problem 96 - Precalculus Mathematics for Calculus

Question

Solve the equation. $$z^{3}-1=0$$

Step-by-Step Solution

Answer

The solutions are \( z = 1 \), \( z = \frac{-1 + i\sqrt{3}}{2} \), and \( z = \frac{-1 - i\sqrt{3}}{2} \).

Step 1: 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.

Step 2: Rewrite the Equation

Recognize \( z^3 - 1 = 0 \) as a difference of cubes. It can be rewritten as \( (z - 1)(z^2 + z + 1) = 0 \).

Step 3: Solve Each Factor Separately

We solve for \( z \) by setting each factor equal to zero: 1. \( z - 1 = 0 \)2. \( z^2 + z + 1 = 0 \).

Step 4: Solve the First Factor

Solve \( z - 1 = 0 \) by adding 1 to both sides, which gives us the solution \( z = 1 \).

Step 5: 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 \( c = 1 \).

Step 6: Apply the Quadratic Formula

Substitute into the quadratic formula: \( z = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times 1}}{2 \times 1} = \frac{-1 \pm \sqrt{-3}}{2} \).

Step 7: 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 \( z = \frac{-1 - i\sqrt{3}}{2} \).

Step 8: List All Solutions

Collect all roots: \( z = 1 \), \( z = \frac{-1 + i\sqrt{3}}{2} \), and \( z = \frac{-1 - i\sqrt{3}}{2} \).

Key Concepts

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:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

The parameters \( a \), \( b \), and \( c \) are coefficients of the equation, and the "\( \pm \)" symbol signifies that there are generally two solutions.
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.

In our original exercise, we used the quadratic formula with \( a = 1 \), \( b = 1 \), and \( c = 1 \) for the second factor \( z^2 + z + 1 = 0 \). Since the discriminant was negative, we found complex solutions, showcasing the importance of this method in both real and complex cases.

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:

• \( z = x + yi \)

Where \( x \) and \( y \) are real numbers. In our example with the quadratic \( z^2 + z + 1 = 0 \), the complex roots found were \( z = \frac{-1 + i\sqrt{3}}{2} \) and \( z = \frac{-1 - i\sqrt{3}}{2} \), each being the conjugate of the other. Complex conjugates are crucial in various mathematical contexts, offering insights into symmetry and behaviors of polynomials in complex planes.

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:

• \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)

This identity breaks a cubic expression into a product of a linear binomial and a quadratic trinomial, which can then be solved separately.
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:

• \( (z - 1)(z^2 + z + 1) = 0 \)

This step simplifies the process, allowing us to set each factor to zero and solve for \( z \). The linear term \( z - 1 = 0 \) yields the real root \( z = 1 \), while the quadratic \( z^2 + z + 1 = 0 \) was addressed using the quadratic formula to find the complex roots. Understanding the difference of cubes is not just essential for solving specific equations but also for grasping deeper algebraic structures and simplifications.