Problem 82 - Precalculus Mathematics for Calculus

Question

Find the indicated roots, and graph the roots in the complex plane. The cube roots of \(4 \sqrt{3}+4 i\)

Step-by-Step Solution

Answer

The cube roots are approximately (1.962+0.348i), (-1.962+0.348i), and (0+2i).

Step 1: Convert to Polar Form

First, convert the complex number to polar form. We have the complex number, \( z = 4 \sqrt{3} + 4i \). The modulus is calculated as \[|z| = \sqrt{(4 \sqrt{3})^2 + (4)^2} = \sqrt{48 + 16} = \sqrt{64} = 8.\]The argument \( \theta \) is given by \[\theta = \tan^{-1}\left(\frac{4}{4\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}.\]Therefore, the polar form of \( z \) is \( 8 \text{cis} \frac{\pi}{6} \) (where cis means \( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \)).

Step 2: Apply the Cube Root Formula

To find the cube roots of a number in polar form, we use De Moivre's Theorem. A cube root of \( z \) is \[z_k = |z|^{1/3} \text{cis} \left(\frac{\theta + 2k\pi}{3}\right)\]for \( k = 0, 1, 2 \). Here, \( |z|^{1/3} = 8^{1/3} = 2 \) and \( \theta = \frac{\pi}{6} \).

Step 3: Find the Roots

Calculate each root:- For \( k = 0 \), \( z_0 = 2 \text{cis} \left(\frac{\frac{\pi}{6} + 0}{3}\right) = 2 \text{cis} \frac{\pi}{18}.\)- For \( k = 1 \), \( z_1 = 2 \text{cis} \left(\frac{\frac{\pi}{6} + 2\pi}{3}\right) = 2 \text{cis} \frac{13\pi}{18}.\)- For \( k = 2 \), \( z_2 = 2 \text{cis} \left(\frac{\frac{\pi}{6} + 4\pi}{3}\right) = 2 \text{cis} \frac{25\pi}{18}.\)

Step 4: Convert Polar to Rectangular Coordinates

Convert each polar form back to rectangular form to plot them in the complex plane.- \( z_0 = 2(\cos \frac{\pi}{18} + i \sin \frac{\pi}{18}) \approx 2 (0.981 + i \cdot 0.174) \approx 1.962 + 0.348i \).- \( z_1 = 2(\cos \frac{13\pi}{18} + i \sin \frac{13\pi}{18}) \approx 2 (-0.981 + i \cdot 0.174) \approx -1.962 + 0.348i \).- \( z_2 = 2(\cos \frac{25\pi}{18} + i \sin \frac{25\pi}{18}) \approx 2 (0 + i \cdot 1) \approx 0 + 2i \).

Step 5: Graph the Roots

Plot the points \( z_0 = 1.962 + 0.348i \), \( z_1 = -1.962 + 0.348i \), and \( z_2 = 0 + 2i \) in the complex plane. These points show the roots evenly spaced on a circle of radius 2.

Key Concepts

Polar Form

Polar form is a way of representing complex numbers using their magnitude (or modulus) and their direction (or angle). This is particularly useful for multiplying and dividing complex numbers or when performing calculations involving powers and roots.
The standard polar form of a complex number is given as \( r \text{cis} \theta \), where:

• \( r \) is the modulus, and it represents the distance from the origin to the point in the complex plane.
• \( \theta \) is the argument, and it is the angle formed with the positive x-axis.

It turns a complex number into a form that is easier to manipulate in certain mathematical operations. For example, the complex number \(4 \sqrt{3} + 4i\) is converted to polar form as \(8 \text{cis} \frac{\pi}{6}\) by finding its modulus \(8\) and angle \(\frac{\pi}{6}\).

Cube Roots

Cube roots of a complex number are solutions to the equation \(z^3 = a\), where \(a\) is the complex number. For any complex number, there are always three cube roots.
To find the cube roots of a complex number in polar form, you:

• Take the cube root of the modulus.
• Divide the angle by 3, and adjust it for each root by adding multiples of \( \frac{2\pi}{3}\), since they need to be spaced evenly around the circle.

So for the polar form \(8 \text{cis} \frac{\pi}{6}\), the cube roots are obtained using these steps, resulting in roots at angles \(\frac{\pi}{18}\), \(\frac{13\pi}{18}\), and \(\frac{25\pi}{18}\).

De Moivre's Theorem

De Moivre's Theorem is a very useful tool for finding powers and roots of complex numbers when expressed in polar form.
This theorem states that if \(z=r\text{cis}\theta\), then for any integer \(n\), \(z^n = r^n \text{cis}(n\theta)\).
For roots, it simplifies the process of finding the roots of complex numbers.

• To find the cube root, for example, you raise the modulus to the power of \(1/3\).
• The angle \(\theta\) is divided by 3, providing each distinct root by adding \(\frac{2\pi k}{3}\) (where \(k\) is a root index).

Applying it to \(8 \text{cis} \frac{\pi}{6}\), we derive the cube roots \(2 \text{cis} \frac{\pi}{18}\), \(2 \text{cis} \frac{13\pi}{18}\), and \(2 \text{cis} \frac{25\pi}{18}\).

Complex Plane

The complex plane is a two-dimensional plane that allows us to graphically represent complex numbers.
In this plane, the horizontal axis (x-axis) represents the real part of complex numbers, while the vertical axis (y-axis) represents the imaginary part.
Plotting in the complex plane, cube roots for \(4 \sqrt{3} + 4i\) look like points on a circle with radius equal to the modulus of the cube root, providing a visual distribution of solutions.

• The cube roots are \(z_0 = 1.962 + 0.348i\).
• Similarly, \(z_1 = -1.962 + 0.348i\).
• And, \(z_2 = 0 + 2i\), all evenly spaced around the center origin.

Rectangular Coordinates

Rectangular coordinates are a way to express complex numbers in the form \(a + bi\), where \(a\) and \(b\) are real numbers.
This form is very intuitive for arithmetic operations and is directly related to the complex plane.

• The real part \(a\) corresponds to the x-coordinate.
• The imaginary part \(b\) corresponds to the y-coordinate.

When converting from polar to rectangular coordinates, the formula \( a = r \cos \theta \) and \( b = r \sin \theta \) is used.
For example, the polar forms were converted as follows:

• \( z_0 \approx 1.962 + 0.348i \)
• \( z_1 \approx -1.962 + 0.348i \)
• \( z_2 \approx 0 + 2i \)

The rectangular form is essential for plotting and interpreting solutions on the complex plane.