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Find every complex number \(z\) such that \(z^{3}=1\). (solution in the pdf version of the book)

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 find all complex numbers \(z\) such that \(z^3 = 1\). This means \(z\) is a root of the equation \(z^3 - 1 = 0\).
02

Factor the Polynomial

The equation \(z^3 - 1 = 0\) can be factored using the difference of cubes formula. Recall that \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). Here, \(a = z\) and \(b = 1\).Thus, we have \(z^3 - 1 = (z - 1)(z^2 + z + 1) = 0\).
03

Solve Each Factor Separately

Set each factor equal to zero:1. \(z - 1 = 0\) leads to \(z = 1\).2. \(z^2 + z + 1 = 0\) is a quadratic equation which we need to solve next.
04

Solve the Quadratic Equation

To solve \(z^2 + z + 1 = 0\), use the quadratic formula: \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 1\), and \(c = 1\).Calculate the discriminant: \(b^2 - 4ac = 1^2 - 4 \times 1 \times 1 = -3\).Since the discriminant is negative, the solutions are complex.
05

Simplify Using Quadratic Formula

Substitute into the quadratic formula:\(z = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}\).This gives us the two complex solutions: \(z = \frac{-1 + i\sqrt{3}}{2}\) and \(z = \frac{-1 - i\sqrt{3}}{2}\).
06

Collect All Solutions

We now have all solutions for \(z^{3} = 1\): 1. \(z = 1\)2. \(z = \frac{-1 + i\sqrt{3}}{2}\)3. \(z = \frac{-1 - i\sqrt{3}}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Roots of Equations
A root of an equation is a number that, when substituted into the equation, makes the equation true. For example, finding the roots of the equation \(z^3 - 1 = 0\) means finding values of \(z\) that make the equation equal to zero. These values are also known as solutions of the equation. In the case of a polynomial equation like \(z^3 - 1 = 0\), there can be multiple roots depending on the degree of the polynomial. For a cubic equation like this one, we typically expect up to three roots.

These roots can be real or complex numbers. A complex number has a real part and an imaginary part, typically written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property that \(i^2 = -1\). Discovering these roots involves both algebraic techniques and some initial guesswork in recognizing possible patterns or standard forms, such as factorization related to known formulas.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). To solve these equations, we use the formula:
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
This method is incredibly useful because it provides a straightforward way to find the roots of any quadratic equation, even when the solutions involve complex numbers. The part under the square root, \(b^2 - 4ac\), is known as the discriminant.

The discriminant indicates the nature of the roots:
  • If \(b^2 - 4ac > 0\), there are two distinct real roots.
  • If \(b^2 - 4ac = 0\), there is one real root, called a repeated or double root.
  • If \(b^2 - 4ac < 0\), the roots are complex and come in a conjugate pair.
In solving the equation \(z^2 + z + 1 = 0\) from the exercise, we found a negative discriminant, leading to complex solutions. The quadratic formula enables us to calculate these outcomes precisely.
Factorization
Factorization refers to expressing a polynomial as a product of simpler polynomials, which can provide immediate insights into the roots of the equation. For the polynomial \(z^3 - 1 = 0\), the method of factorization was used by applying the difference of cubes formula:
  • \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)
In this context, we let \(a = z\) and \(b = 1\), legally breaking it down into a product of the linear factor \((z - 1)\) and the quadratic factor \((z^2 + z + 1)\).

Such factorization is crucial as it lays the foundation for more straightforward equations to solve. Each factor potentially represents a root of the original equation. The cubic equation \(z^3 - 1 = 0\) is thus turned into two problems:
  • Simpler linear equation: \(z - 1 = 0\)
  • Solvable quadratic equation: \(z^2 + z + 1 = 0\)
This allows us to systematically approach and determine each root, whether real or complex.
Difference of Cubes
The difference of cubes formula is a particular algebraic identity that allows us to factor expressions of the form \(a^3 - b^3\). It is a specific case of the broader category of factorization identities and is expressed as:
  • \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)
This identity helps break down seemingly complicated cubic expressions into manageable parts. By applying this formula, we were able to transform the cubic equation \(z^3 - 1 = 0\) into a simpler expression with its factors.

Understanding how to utilize this formula is vital in algebra because it provides an efficient way to solve otherwise complex equations. Recognizing the structure of \(a^3 - b^3\) allows us to identify and handle specific types of polynomial equations, unlocking the solutions to equations that may initially seem daunting. This approach showcases how algebraic identities underpin many solutions in higher mathematics.

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