Problem 58 - A Graphical Approach to College Algebra

Question

Find all \(n\) complex solutions of each equation of the form \(x^{n}=k\) $$x^{3}=-8$$

Step-by-Step Solution

Answer

The roots are \(1+i\sqrt{3}, -2, 1-i\sqrt{3}\).

Step 1:Identify the Form of the Equation

The given equation is \(x^3 = -8\). The general form for such equations is \(x^{n} = k\). Here, \(n = 3\) and \(k = -8\). We are tasked with finding all the complex solutions for \(n = 3\).

Step 2:Convert to Polar Form

The solution involves converting \(k = -8\) to its polar form. To do this, express \(-8\) as a complex number: \(-8 = 8(\cos \pi + i\sin \pi)\), which in polar form is \(8\text{cis}(\pi)\) where \(\text{cis}(\theta) = \cos(\theta) + i\sin(\theta)\).

Step 3:Apply De Moivre's Theorem

Using De Moivre's Theorem, we find the cube roots of \(8\text{cis}(\pi)\). The formula for the \(n\)th roots is \(r^{1/n} \text{cis}((\theta + 2k\pi)/n)\), for \(k = 0, 1, ..., n-1\).

Step 4:Calculate the Roots Using the Formula

We calculate three roots since \(n = 3\). The modulus of the root is \(8^{1/3} = 2\).- For \(k = 0\): \(z_0 = 2 \text{cis}(\pi/3) = 2(\cos(\pi/3) + i\sin(\pi/3)) = 2(1/2 + i\sqrt{3}/2) = 1 + i\sqrt{3}\).- For \(k = 1\): \(z_1 = 2 \text{cis}(\pi + 2\pi/3) = 2 \text{cis}(\pi) = 2(-1 + i0) = -2\).- For \(k = 2\): \(z_2 = 2 \text{cis}(5\pi/3) = 2(\cos(5\pi/3) + i\sin(5\pi/3)) = 2(1/2 - i\sqrt{3}/2) = 1 - i\sqrt{3}\).

Step 5:List the Solutions

The three complex roots of the equation \(x^3 = -8\) are: \(1 + i\sqrt{3}\), \(-2\), and \(1 - i\sqrt{3}\).

Key Concepts

De Moivre's Theorem

De Moivre's Theorem is a powerful tool for working with complex numbers, especially when they are in polar form. It provides a method for raising complex numbers to a power or extracting roots. The theorem states that for any complex number in polar form \( r \text{cis}(\theta) \) and integer \( n \), we have:\[(r \text{cis}(\theta))^n = r^n \text{cis}(n\theta)\]When dealing with roots, we apply the inverse of raising to a power. If you need to find the nth roots of a complex number, you can use:\[r^{1/n} \text{cis}\left( \frac{\theta + 2k\pi}{n} \right)\]Here, \( k = 0, 1, ..., n-1 \). This process divides the angle equally for each root around the circle, effectively distributing the roots uniformly. Each resulting root is one of the solutions to the original complex equation. This method is especially useful as it allows us to simply and systematically find all possible roots of complex equations.

Polar Form

The polar form of a complex number offers a different way of representing it as compared to the standard Cartesian form (a + bi). Instead of using rectangular coordinates, complex numbers in polar form are expressed using a magnitude (or modulus) and an angle (or argument). A complex number \( a + bi \) can be converted into polar form:\[r(\cos(\theta) + i\sin(\theta))\]This can be further simplified into the shorthand notation:\[r \text{cis}(\theta)\]Where \( r \) is the magnitude and \( \theta \) is the angle measured from the positive x-axis.

• The magnitude \( r \) is calculated as \( \sqrt{a^2 + b^2} \).
• The angle \( \theta \), or argument, is found using \( \tan^{-1}(b/a) \).

This method simplifies the multiplication and division of complex numbers, making polar form very practical in problems involving powers and roots. It's particularly necessary when applying De Moivre's Theorem.

Nth Roots of Complex Numbers

To find the nth roots of a complex number, especially in the context of equations like \( x^n = k \), you'll often employ De Moivre's Theorem as discussed above. Here's how the process works:1. **Convert to Polar Form:** First, express the complex number (say \( k \)) in its polar form as \( r \text{cis}(\theta) \).2. **Use the Root Formula:** Apply the formula \( r^{1/n} \text{cis}\left( \frac{\theta + 2k\pi}{n} \right) \), where \( k = 0, 1, ..., n-1 \).

• Each value of \( k \) generates a different nth root.

3. **Calculate the Results:** Substitute the values in the formula to find each root. You will have \( n \) distinct solutions for \( n \) roots, covering all possible solutions.This approach uses the full representation of a complex circle, spaced evenly in angles by \( 2\pi/n \), providing all potential candidates for the roots. This methodology is fundamental in complex analysis and is a vital concept for solving polynomial equations with complex coefficients.