Problem 87 - 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 cube roots of \(\frac{27 \sqrt{2}}{2}+\frac{27 \sqrt{2}}{2} i,\) and draw the complex triangle with a calculator.

Step-by-Step Solution

Answer

The cube roots of the complex number are given by 3 times the cube roots of unity at angles \( \frac{\pi}{12}, \frac{9\pi}{12}, \text{and} \frac{17\pi}{12} \).

Step 1: Convert to Polar Form

First, convert the complex number \[ z = \frac{27 \sqrt{2}}{2} + \frac{27 \sqrt{2}}{2} i \]into polar form. Calculate the modulus \( r \) and argument \( \theta \).The modulus \( r \) is given by \[ r = \sqrt{ \left(\frac{27 \sqrt{2}}{2}\right)^2 + \left(\frac{27 \sqrt{2}}{2}\right)^2 } = 27. \]The argument \( \theta \) is \[ \theta = \tan^{-1} \left( \frac{ \frac{27 \sqrt{2}}{2} }{ \frac{27 \sqrt{2}}{2} } \right) = \frac{\pi}{4} \text{ radians}. \]Therefore, in polar form, \[ z = 27 \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right). \]

Step 2: Find Cube Roots

Apply the nth root theorem to find the cube roots.The first cube root is \[ w_1 = 27^{1/3} \left[ \cos \left( \frac{\frac{\pi}{4}}{3} + \frac{2\pi \times 0}{3} \right) + i \sin \left( \frac{\frac{\pi}{4}}{3} + \frac{2\pi \times 0}{3} \right) \right]. \]Calculate \( 27^{1/3} = 3 \) and simplify the angles:\[ w_1 = 3 \left[ \cos \frac{\pi}{12} + i \sin \frac{\pi}{12} \right]. \]Repeat for the next two roots:- Second root: \( w_2 = 3 \left[ \cos \left( \frac{\pi}{12} + \frac{2\pi}{3} \right) + i \sin \left( \frac{\pi}{12} + \frac{2\pi}{3} \right) \right]. \)- Third root: \( w_3 = 3 \left[ \cos \left( \frac{\pi}{12} + \frac{4\pi}{3} \right) + i \sin \left( \frac{\pi}{12} + \frac{4\pi}{3} \right) \right]. \)

Step 3: Plot the Cube Roots

Use a graphing calculator to plot the cube roots by entering the appropriate settings.- Set \( r = 3 \), \( \theta \) min as \( \frac{\pi}{12} \) and \( \theta \) max as \( \frac{25\pi}{12} \) with \( \theta \) step \( \frac{2\pi}{3} \).- Set \( \text{xmin} = -3 \), \( \text{xmax} = 3 \), \( \text{ymin} = -3 \), \( \text{ymax} = 3 \).- Ensure the calculator is in radian mode.Plot each \( w_k \) using these polar coordinates to form a triangle, representing the cube roots.

Key Concepts

Complex Numbers

Complex numbers are a fascinating extension of the real numbers, incorporating the element "i", known as the imaginary unit. The imaginary unit is defined such that \( i^2 = -1 \), which leads to every complex number being expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers. - The part \( a \) is called the real component.- The part \( bi \) is called the imaginary component.The beauty of complex numbers lies in their utility in representing points in a two-dimensional space, making them very effective for various mathematical calculations and applications. They form ordered pairs on the complex plane, resembling Cartesian coordinates but with enhanced capabilities. In this exercise, the complex number given is \( \frac{27 \sqrt{2}}{2} + \frac{27 \sqrt{2}}{2}i \), which shows equal real and imaginary parts. Such numbers align along a diagonal line from the origin in the complex plane.

Polar Coordinates

Polar coordinates are a system where a point in the plane is determined by a distance from a reference point and an angle from a reference direction. - The reference point is called the pole, usually the origin, and the direction is the positive x-axis.- A point is described in terms of the modulus, \( r \), representing its distance from the origin, and the argument, \( \theta \), the angle from the positive x-axis.Converting a complex number to polar form is crucial for solving many problems, like finding roots. To express the number \( z = 27 \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) \) in polar coordinates, we identify \( r = 27 \), and the angle \( \theta = \frac{\pi}{4} \) radians. This form is particularly beneficial because it allows complex multiplication and division to be simply implemented using trigonometric relationships, making operations with powers and roots more intuitive.

Cube Roots

Finding the cube roots of a complex number involves using the n-th Root Theorem. This theorem provides a formula to calculate the cube roots. Cube roots of a number are the three values that, when cubed, give the original number. The steps involve:1. **Finding the Modulus:** Based on the given complex number, we start by determining the modulus, calculated as 27. The cube root of 27 is 3.2. **Calculating the Argument:** Next, divide the argument by 3. The original angle \( \theta = \frac{\pi}{4} \) divided by 3 is \( \frac{\pi}{12} \).The first cube root \( w_1 \) becomes \( 3 \left( \cos \frac{\pi}{12} + i \sin \frac{\pi}{12} \right) \).Repeat the calculation by adding \( \frac{2 \pi}{3} \) for the second root and \( \frac{4 \pi}{3} \) for the third. This cyclical addition accounts for the symmetrical nature of roots around the origin in the complex plane. So, the angles for the subsequent roots illustrate their equal spacing on a circle, reminiscent of vertices of an equilateral triangle on the complex plane.