Problem 85 - Precalculus

Question

Refer to the following: According to the \(n\) th root theorem, the first of the \(n\) th roots of the complex number \(z=r(\cos \theta+i \sin \theta)\) is given by \(w_{1}=r^{1 / n}\left[\cos \left(\frac{\theta}{n}+\frac{2 \pi}{n}\right)+i \sin \left(\frac{\theta}{n}+\frac{2 \pi}{n}\right)\right],\) with \(\theta\) in radians or \(w_{1}=r^{1 / n}\left[\cos \left(\frac{\theta}{n}+\frac{360^{\circ}}{n}\right)+i \sin \left(\frac{\theta}{n}+\frac{360^{\circ}}{n}\right)\right]\) with \(\theta\) in degrees. Using the graphing calculator to plot the \(n\) roots of a complex number \(z,\) enter \(r_{1}=r, \theta \min =\frac{\theta}{n}, \theta \max =2 \pi+\frac{\theta}{n}\) or \(360^{\circ}+\frac{\theta}{n}, \theta\) step \(=\frac{2 \pi}{n}\) or \(\frac{360^{\circ}}{n}, \mathrm{xmin}=-r, \mathrm{xmax}=r\) ymin \(=-r,\) ymax \(=r,\) and \([\mathrm{MODE}]\) in radians or degrees. Find the sixth roots of \(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\), and draw the complex hexagon with a calculator.

Step-by-Step Solution

Answer

The sixth roots of \\(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\\) form a hexagon inscribed in the unit circle.

Step 1: Convert to Polar Form

First, convert the complex number \(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\) to its polar form \((r(\cos \theta + i \sin \theta))\). Calculate the modulus \(r\) using the formula \[ r = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]Next, find the angle \(\theta\) using the tangent inverse:\[ \theta = \tan^{-1}\left(\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\right) = \tan^{-1}(\sqrt{3}) = \frac{2\pi}{3} \text{ radians} \].Hence, the polar form is \(z = 1(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})\).

Step 2: Calculate First Root

Use the \(-th\) root theorem to find the first root.Set \(n = 6\) (since we're finding sixth roots):\[ w_1 = 1^{1/6} \left[ \cos\left(\frac{\frac{2\pi}{3}}{6} + \frac{2\pi \times 0}{6}\right) + i \sin\left(\frac{\frac{2\pi}{3}}{6} + \frac{2\pi \times 0}{6}\right) \right] \]\[ w_1 = \cos\left(\frac{\pi}{9}\right) + i \sin\left(\frac{\pi}{9}\right) \]

Step 3: Find All Sixth Roots

Calculate the remaining sixth roots by rotating by \(\frac{2\pi}{6}\) radians each time. Define the general formula for the roots as:\[ w_k = \cos\left(\frac{\pi}{9} + k\frac{2\pi}{6}\right) + i \sin\left(\frac{\pi}{9} + k\frac{2\pi}{6}\right) \] where \(k = 0, 1, 2, 3, 4, 5\).Instead of manually calculating each angle, think of this process as creating a full circle by dividing it into \(6\) equal parts, each adding \(\frac{\pi}{3}\) radians to the previous.

Step 4: Plot on a Graphing Calculator

Using a graphing calculator, set up the plots for the roots.- \(r_1 = 1\)- \(\theta \min = \frac{\pi}{9}, \theta \max = 2\pi + \frac{\pi}{9}\)- \(\theta \text{ step} = \frac{2\pi}{6} = \frac{\pi}{3}\)- \(\text{xmin} = -1, \text{xmax} = 1\)- \(\text{ymin} = -1, \text{ymax} = 1\)- Set mode to radians.Plot these points to see the vertices of a hexagon formed by the sixth roots.

Key Concepts

Understanding Polar Form Conversion

The Polar Form Conversion is a method used to express complex numbers. A complex number such as \(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\) can be converted into its polar form \(r(\cos \theta + i \sin \theta)\). This conversion is useful because it simplifies the multiplication and division of complex numbers. To convert to polar form, first, calculate the modulus \(r\) using the formula \( r = \sqrt{a^2 + b^2} \), where \(a\) and \(b\) are the real and imaginary parts respectively. In this case, the modulus is 1. Next, find the angle \(\theta\) using the tangent inverse of the ratio between the imaginary and real parts. Here, \(\theta = \tan^{-1}(\sqrt{3}) = \frac{2\pi}{3}\) in radians. The polar form thus becomes \(z = 1(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})\). This compact expression now represents the complex number's magnitude and direction efficiently.

Exploring the Nth Root Theorem

The Nth Root Theorem is a valuable tool for finding the roots of complex numbers. According to this theorem, the \(n\) different roots of a complex number can be found by referencing its polar form. Each root can be expressed as \( w_k = r^{1/n} \left[ \cos \left(\frac{\theta}{n} + k\frac{2\pi}{n}\right) + i \sin \left(\frac{\theta}{n} + k\frac{2\pi}{n}\right) \right] \), where \(k\) spans from 0 to \(n-1\). These roots form a regular polygon on the complex plane.
For example, to find the sixth roots of a complex number like \(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\), set \(n = 6\). Each root shares the same modulus, \(r^{1/n} = 1^{1/6} = 1\), since the modulus of the original number is 1. Calculate the angles for these roots by adding \(\frac{2\pi}{6}\), or \(\frac{\pi}{3}\), starting from the base angle \(\frac{\pi}{9}\). With each increment, a different root angle is calculated, spacing these roots evenly along the unit circle and forming a hexagon.

Graphing Calculator Techniques

Graphing calculators are excellent tools for visualizing the roots of complex numbers. When you're tasked with plotting the sixth roots of a number like \(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\), follow these setup steps to illustrate the results accurately.

• Set the radius, \(r_1 = 1\), reflecting the modulus of our complex number.
• Define angle limits: \(\theta \min = \frac{\pi}{9}\) and \(\theta \max = 2\pi + \frac{\pi}{9}\).
• Specify the angle step: \(\theta \text{ step} = \frac{2\pi}{6} = \frac{\pi}{3}\), ensuring the calculated angles sweep through the full circle.
• Adjust viewing windows to encompass the potential coordinates: \(\text{xmin} = -1, \text{xmax} = 1\) and \(\text{ymin} = -1, \text{ymax} = 1\).
• Make sure the calculator is in radians mode for consistency.

With these settings, the graphing calculator will plot six equidistant points along a circle's circumference, forming a hexagon that represents the sixth roots visually, combining computational power with geometric insight.