Problem 51 - A Graphical Approach to Precalculus with Limits

Question

Find all complex solutions for each equation. Leave your answers in trigonometric form. $$x^{3}-8=0$$

Step-by-Step Solution

Answer

The complex solutions are \( 2 \text{cis} 0 \), \( 2 \text{cis} \frac{2\pi}{3} \), and \( 2 \text{cis} \frac{4\pi}{3} \).

Step 1: Rewrite the Equation

Start by rewriting the equation in the form of a difference of cubes: \[ x^3 - 2^3 = 0 \]

Step 2: Apply the Difference of Cubes Formula

The difference of cubes formula is \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Applying this formula:\[ x^3 - 2^3 = (x - 2)(x^2 + 2x + 4) = 0 \]

Step 3: Solve the Linear Factor

The first factor, \( x - 2 = 0 \), gives the real solution:\[ x = 2 \] which can be expressed in trigonometric form as \( 2 \text{cis} 0 \).

Step 4: Solve the Quadratic Equation

The quadratic equation is \( x^2 + 2x + 4 = 0 \). Use the quadratic formula to find the complex solutions:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substituting \( a = 1, b = 2, c = 4 \):\[ x = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2} \]\[ x = \frac{-2 \pm 2\sqrt{3}i}{2} \]\[ x = -1 \pm \sqrt{3}i \]

Step 5: Express Complex Solutions in Trigonometric Form

Convert \( x = -1 + \sqrt{3}i \) and \( x = -1 - \sqrt{3}i \) to trigonometric form:Calculate the modulus:\[ r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 \]For \( -1 + \sqrt{3}i \):\[ \theta = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right) = \frac{2\pi}{3} \]Trigonometric form: \[ 2 \text{cis} \frac{2\pi}{3} \]For \( -1 - \sqrt{3}i \):\[ \theta = \tan^{-1}\left(\frac{-\sqrt{3}}{-1}\right) = \frac{4\pi}{3} \]Trigonometric form: \[ 2 \text{cis} \frac{4\pi}{3} \]

Key Concepts

Understanding the Difference of Cubes

The expression \(x^3 - 8\) can be rewritten as \(x^3 - 2^3\), which is a classic example of a difference of cubes. This is a helpful algebraic identity as it allows us to factorize expressions more easily.
The difference of cubes formula is \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). This means any expression in this form can be split into two factors. The first is a simple linear term \((a - b)\), and the second is a quadratic \((a^2 + ab + b^2)\).
By applying this formula to \(x^3 - 2^3\), we express it as \((x - 2)(x^2 + 2x + 4)\). This factorization is crucial in solving the equation since it simplifies the process by breaking it down into easier parts.

Solving the Quadratic Equation

After using the difference of cubes, we end up with the equation \((x - 2)(x^2 + 2x + 4) = 0\). The factor \(x - 2\) gives us a straightforward real solution, \(x = 2\). However, the quadratic \(x^2 + 2x + 4 = 0\) needs further solving.
To solve it, we use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Here, \(a = 1\), \(b = 2\), and \(c = 4\). Plugging these values in:

• \(b^2 - 4ac = 4 - 16 = -12\)
• The solutions are \(x = \frac{-2 \pm \sqrt{-12}}{2}\)
• Simplifying gives \(x = -1 \pm \sqrt{3}i\)

These solutions will lead us to the complex solutions of the original equation.

Complex Solutions in Trigonometric Form

Complex numbers can be converted into trigonometric form to simplify multiplication and division. The trigonometric form of a complex number is \(r \text{cis} \theta\), where \(r\) is the modulus and \(\theta\) is the argument.
For the complex solutions \(-1 + \sqrt{3}i\) and \(-1 - \sqrt{3}i\), let's calculate:

• **Modulus (r):** \[r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{4} = 2\]
• **Argument (\(\theta\)) for \(-1 + \sqrt{3}i\):** \[\theta = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right) = \frac{2\pi}{3}\]
• **Trigonometric form:** \(2 \text{cis} \frac{2\pi}{3}\)
• **Argument (\(\theta\)) for \(-1 - \sqrt{3}i\):** \[\theta = \tan^{-1}\left(\frac{-\sqrt{3}}{-1}\right) = \frac{4\pi}{3}\]
• **Trigonometric form:** \(2 \text{cis} \frac{4\pi}{3}\)

Modulus and Argument of Complex Numbers

The modulus and argument are key components in expressing complex numbers in trigonometric form. The modulus, \(r\), represents the distance from the origin to the point in the complex plane. It is calculated using \(r = \sqrt{x^2 + y^2}\), where \(x\) and \(y\) are the real and imaginary parts.
The argument, \(\theta\), is the angle formed with the positive x-axis. It can be found using the inverse tangent function: \(\theta = \tan^{-1}(\frac{y}{x})\). When placing the complex number in trigonometric form, the angle needs to be in the correct quadrant:

• First quadrant: Angle is directly \(\theta\)
• Second quadrant: Angle is \(\pi - \theta\)
• Third quadrant: Angle is \(\pi + \theta\)
• Fourth quadrant: Angle is \(2\pi - \theta\)

Understanding this helps in expressing any complex number in trigonometric form, which is especially useful in simplifying calculations.