Problem 58 - A Graphical Approach to Precalculus with Limits

Question

We examine how the three complex cube roots of \(-8\) can be found in two different ways. Use the method described in this section to find the three complex cube roots of \(-8 .\) Give them in trigonometric form.

Step-by-Step Solution

Answer

The complex cube roots of \(-8\) are \(2\text{cis}\left(\frac{\pi}{3}\right)\), \(-2\), and \(2\text{cis}\left(\frac{5\pi}{3}\right)\).

Step 1: Determine Modulus and Argument

The cube roots of a complex number can be found by expressing the number in trigonometric form. First, find the modulus: \(|-8| = 8\), because it is the absolute value of the real number. The argument (angle) of \(-8\) in the complex plane is, by definition, \(\pi\) because \(-8\) lies on the negative real axis.

Step 2: Express in Polar Form

Express \(-8\) as a complex number in polar form: \(-8 = 8\text{cis}(\pi)\), where \(\text{cis}(\theta) = \cos(\theta) + i\sin(\theta)\).

Step 3: Apply De Moivre's Theorem

To find the cube roots of \(-8\), apply De Moivre's Theorem: \(z = r^{1/n} \text{cis}\left(\frac{\theta + 2k\pi}{n}\right)\) for \(k = 0, 1, 2\) and \(n = 3\). Here, \(r = 8\), \(\theta = \pi\), and \(n = 3\). This gives \(z = 2\text{cis}\left(\frac{\pi + 2k\pi}{3}\right)\).

Step 4: Compute Cube Roots for k = 0

For \(k = 0\), compute the root: \(z = 2\text{cis}\left(\frac{\pi}{3}\right) = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\).

Step 5: Compute Cube Roots for k = 1

For \(k = 1\), compute the root: \(z = 2\text{cis}\left(\frac{\pi + 2\pi}{3}\right) = 2\text{cis}\left(\pi\right) = 2(-1 + 0i) = -2\).

Step 6: Compute Cube Roots for k = 2

For \(k = 2\), compute the root: \(z = 2\text{cis}\left(\frac{\pi + 4\pi}{3}\right) = 2\text{cis}\left(\frac{5\pi}{3}\right)\).

Step 7: List the Cube Roots in Trigonometric Form

The three complex cube roots of \(-8\) in trigonometric form are: \(2\text{cis}\left(\frac{\pi}{3}\right)\), \(-2\), and \(2\text{cis}\left(\frac{5\pi}{3}\right)\).

Key Concepts

Cube Roots

Finding the cube roots of a number is about discovering which numbers can be multiplied by themselves three times to get the original number. When dealing with complex numbers, this process involves finding not just one, but several roots. This is because the cube root of a complex number results in three different numbers due to the circular nature of complex numbers on the complex plane. The exercise seeks to find the complex cube roots of \(-8\).
\(-8\) can be considered as a point on the complex plane, specifically on the negative real axis. Since we are working with cube roots, we can understand that we will find three equally spaced roots around the origin, forming a shape reminiscent of an equilateral triangle when visualized. This perfectly illustrates the symmetry and beauty inherent in complex numbers and their representations.
In our solution, we use both the trigonometric and polar forms to efficiently find these roots, emphasizing the convenience and practicality of expressing complex numbers in these forms for calculations involving powers or roots.

Trigonometric Form

Expressing a complex number in trigonometric form involves representing it using a modulus and an angle, or argument. This form is particularly useful for complex number calculations such as finding roots or powers.
In trigonometric form, a complex number is expressed as \( r \text{cis}(\theta) \), where \( r \) is the modulus (the distance from the origin in the complex plane) and \( \theta \) is the angle the number makes with the positive real axis. The expression \( \text{cis}(\theta) \) is shorthand for \( \cos(\theta) + i\sin(\theta) \).
For the number \(-8\), which lies on the negative real axis, the modulus is 8, and the argument is \( \pi \) since it's exactly opposite the positive real axis. Therefore, in trigonometric form, \(-8\) is represented as \( 8 \text{cis}(\pi) \). This trigonometric representation simplifies the process of finding the cube roots using De Moivre's Theorem later on.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool for simplifying the process of finding powers and roots of complex numbers in trigonometric form. The theorem states that if a complex number is in the form \( r \text{cis}(\theta) \), then its \( n \)th power is \( r^n \text{cis}(n\theta) \), and similarly, its \( n \)th root is \( r^{1/n} \text{cis}\left(\frac{\theta + 2k\pi}{n}\right) \). Here, \( k \) represents different integers to ensure all roots are found.
In the case of \(-8\) and its cube roots, \r = 8\, and \theta = \pi\. Using De Moivre’s Theorem for\( n = 3 \), the cube roots are found as follows:

• For \( k = 0 \), the root is \( 2 \text{cis}\left(\frac{\pi}{3}\right) \).
• For \( k = 1 \), the root results in \( -2 \), simplifying the polar representation \( 2 \text{cis}(\pi) \).
• For \( k = 2 \), the root is \( 2 \text{cis}\left(\frac{5\pi}{3}\right) \).

De Moivre's Theorem accommodates the cyclical and periodic nature of complex numbers and angles, facilitating the computation of roots like these efficiently and elegantly.