Problem 28 - A Graphical Approach to Precalculus with Limits

Question

Find the cube roots of each complex number. Leave the answers in trigonometric form. Then graph each cube root as a vector in the complex plane. $$2-2 i \sqrt{3}$$

Step-by-Step Solution

Answer

The cube roots are: \(1.5874\angle -\pi/9\), \(1.5874\angle \pi/3\), \(1.5874\angle \pi\).

Step 1: Convert to Polar Form

First, express the complex number \(2 - 2i \sqrt{3}\) in polar form. Calculate the modulus: \(r = \sqrt{2^2 + (-2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4\). Calculate the argument \( \theta \) using \( \tan^{-1}\left(\frac{-2\sqrt{3}}{2}\right) = \tan^{-1}(-\sqrt{3})\). This corresponds to an angle \( -\frac{\pi}{3} \). Thus, the polar form is \(4(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3}))\).

Step 2: Find the Principal Cube Root

The cube root will have the form \( \sqrt[3]{4}\left(\cos\left(\frac{-\pi}{3} + \frac{2k\pi}{3}\right) + i\sin\left(\frac{-\pi}{3} + \frac{2k\pi}{3}\right)\right)\) where \( k = 0, 1, 2 \).Start with \( k = 0 \): \( \sqrt[3]{4} = 4^{1/3} \approx 1.5874 \). The angle becomes \( \frac{-\pi}{9} \). Thus, the first root is \(1.5874(\cos(-\pi/9) + i \sin(-\pi/9))\).

Step 3: Find the Other Two Cube Roots

Use the same formula for \( k = 1 \) and \( k = 2 \): - For \( k = 1\), the angle is \(\frac{-\pi}{3} + \frac{2\pi}{3} = \frac{\pi}{3}\). This gives the second root: \(1.5874(\cos(\pi/3) + i \sin(\pi/3))\). - For \( k = 2 \), the angle is \(\frac{-\pi}{3} + \frac{4\pi}{3} = \pi\). This gives the third root: \(1.5874(\cos(\pi) + i \sin(\pi))\).

Step 4: Express the Roots in Trigonometric Form

Summarize the roots:1. \(1.5874(\cos(-\pi/9) + i \sin(-\pi/9))\)2. \(1.5874(\cos(\pi/3) + i \sin(\pi/3))\)3. \(1.5874(\cos(\pi) + i \sin(\pi))\)These are the three cube roots in trigonometric form.

Step 5: Graph Each Root as a Vector

Graph each root on the complex plane as a vector starting from the origin. - The first vector corresponds to the angle \(-\pi/9\) with magnitude \(1.5874\). - The second vector corresponds to the angle \(\pi/3\) with magnitude \(1.5874\). - The third vector corresponds to the angle \(\pi\) with magnitude \(1.5874\).This setups an equilateral triangle centered at the origin with vertices corresponding to the roots.

Key Concepts

Trigonometric Form

The trigonometric form of a complex number offers a convenient way to handle complex arithmetic operations, especially when dealing with powers and roots. This form leverages the concept of expressing a complex number in terms of cosine and sine functions. For a given complex number,
the trigonometric form is represented as \(r(\cos \theta + i\sin \theta)\),
where \(r\) is the modulus (or absolute value) and \(\theta\) is the argument (or angle).
The use of trigonometric form simplifies operations like finding roots, as demonstrated with the given exercise. By converting the number into trigonometric form, we are able to directly apply De Moivre's Theorem,
which states that the \(n\)th root of a complex number can be found by extracting the \(n\)th root of its modulus and dividing its argument by \(n\).
This makes solving problems involving cube roots, as shown here, more intuitive and less labor-intensive.

Complex Plane

The complex plane is a vital graphical tool for visualizing complex numbers and their roots.
Viewed as an extension of the Cartesian plane, it establishes a system where the horizontal axis represents the real part of a complex number and the vertical axis represents the imaginary part.
In this plane, every complex number correlates to a distinct point or vector.
When dealing with root problems, such as finding cube roots, each root corresponds to a specific point in this plane, and these points often form geometrical patterns.

• Vectors plotted originate from the origin.
• The angle the vector makes with the positive x-axis is critical, as it represents the argument of the number.
• The length of the vector, or its modulus, is also key to its position.

For the exercise at hand, plotting the cube roots forms an equilateral triangle, illustrating the symmetry and uniform distribution of roots in the complex plane.

Polar Form

Polar form is an alternative representation of complex numbers based on the polar coordinate system. It emphasizes the geometrical viewpoint by focusing on magnitude and direction rather than real and imaginary components.

The transformation from the rectangular to the polar form usually involves calculating the modulus \(r\) and the argument \(\theta\), providing a means to express complex numbers as \(r\angle \theta\) or equivalently \(r(\cos \theta + i\sin \theta)\).
The exercise begins with converting the number \(2 - 2i\sqrt{3}\) into its polar form. This conversion enables easier calculations for the cube roots because the operations become straightforward, leveraging the polar form's strengths.

• The modulus is the distance of the complex number from the origin in the complex plane.
• The argument is the angle it makes with the positive real axis, measured in radians or degrees.

Through polar conversion, operations such as multiplication, division, and finding roots become elegantly simple, exploiting the nature of exponential functions linked with cosine and sine through Euler's formula.