Problem 96 - Algebra and Trigonometry

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: Factor the equation

The given equation is \( z^3 - 1 = 0 \). Recognize this as a difference of cubes which can be factored using the formula \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). Here, \( a = z \) and \( b = 1 \). Thus, the equation can be factored as: \( (z-1)(z^2 + z + 1) = 0 \).

Step 2: Solve the linear factor

Solve the linear factor \( z - 1 = 0 \) to find one solution. \( z = 1 \) is a solution.

Step 3: Solve the quadratic factor

The remaining factor from Step 1 is \( z^2 + z + 1 = 0 \). We solve this quadratic equation using the quadratic formula: \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = 1 \).

Step 4: Calculate the discriminant

Calculate the discriminant of the quadratic, \( b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \). Since the discriminant is negative, the solutions will be complex numbers.

Step 5: Find the complex roots

Using the quadratic formula for \( z^2 + z + 1 = 0 \), substitute the values: \( z = \frac{-1 \pm \sqrt{-3}}{2} \). Simplify this result to find the complex roots: \( z = \frac{-1 \pm i \sqrt{3}}{2} \). Therefore, the complex solutions are \( z = \frac{-1 + i \sqrt{3}}{2} \) and \( z = \frac{-1 - i \sqrt{3}}{2} \).

Key Concepts

Difference of Cubes

Algebra includes many useful techniques, one of them is factoring expressions like the difference of cubes. The expression \(a^3 - b^3\) can be rewritten using the formula:

• \((a-b)(a^2 + ab + b^2)\)
• This formula is especially handy for breaking down higher power polynomials.

Recognizing this structure in an equation makes solving it more manageable. In the original exercise, we have \(z^3 - 1 = 0\), which matches the difference of cubes pattern where \(a = z\) and \(b = 1\). Using the formula, it factors into:

• \((z - 1)(z^2 + z + 1) = 0\)
• Factoring like this separates the equation into parts we can solve individually for their roots.

This step is a significant simplification, leaving us with one linear equation and one quadratic equation to solve.

Quadratic Formula

The quadratic formula is a powerful tool used for solving quadratic equations of the form \(ax^2 + bx + c = 0\). The solutions can be calculated with:

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

Using the quadratic formula requires plugging in the coefficients \(a\), \(b\), and \(c\) from the equation. In our case, dealing with \(z^2 + z + 1 = 0\), we identify:

• \(a = 1\), \(b = 1\), and \(c = 1\).
• Substituting these into the formula provides us the values for finding \(z\).

The ability to use this formula allows solving any quadratic equation, regardless of its complexity or coefficients.

Discriminant

The discriminant is a crucial part of understanding what type of solutions a quadratic equation may have. The discriminant is found by calculating \(b^2 - 4ac\) in the quadratic formula.

• If the discriminant is positive, there will be two distinct real solutions.
• If zero, there is exactly one real solution.
• If negative, it indicates two complex solutions.

This helps gauge what kinds of numbers we expect as solutions. For the equation \(z^2 + z + 1 = 0\), we find the discriminant as:

• \(b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = -3\)
• The negative discriminant shows the roots will be complex numbers.

Understanding discriminants enables assessing the nature of solutions without solving the whole equation.