/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 82 Solve each equation in the compl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 82

Solve each equation in the complex number system. Express solutions in polar and rectangular form. $$x^{6}+1=0$$

Short Answer

Expert verified
The six solutions in the complex number system (in rectangular form) are: \(x_{0} = \frac{\sqrt{3}}{2} + \frac{i}{2}\), \(x_{1} = i\), \(x_{2} = -\frac{\sqrt{3}}{2} + \frac{i}{2}\), \(x_{3} = -\frac{\sqrt{3}}{2} - \frac{i}{2}\), \(x_{4} = -i\), and \(x_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}\). They can also be expressed in polar form.

Step by step solution

01

Rearrange the equation

To find the solution, the equation is first rearranged to \(x^{6} = -1\)
02

Find the solution using De Moivre's theorem

By using De Moivre's theorem, it is known that the solutions of \(x^{n} = cos(胃) + isin(胃)\) are given by \(x_{k} = \sqrt[n]{r}( cos(\frac{胃+2k蟺}{n}) + isin(\frac{胃+2k蟺}{n})), (k = 0, 1, 路路路 , n-1)\). Here, \(r=1\) (the magnitude of -1), \(胃=蟺\) (the argument of -1), \(n=6\) (the exponent of x). Substituting these values gives \(x_{k} = cos(\frac{蟺+2k蟺}{6}) + isin(\frac{蟺+2k蟺}{6}\), for \(k = 0, 1, ... , 5\).
03

Calculate specific solutions

Substituting each value of k in the equation from the previous step produces the following six solutions:\n\n\(x_{0} = cos(\frac{蟺}{6}) + isin(\frac{蟺}{6})\),\n\n\(x_{1} = cos(\frac{蟺}{2}) + isin(\frac{蟺}{2})\),\n\n\(x_{2} = cos(\frac{5蟺}{6}) + isin(\frac{5蟺}{6})\),\n\n\(x_{3} = cos(\frac{7蟺}{6}) + isin(\frac{7蟺}{6})\),\n\n\(x_{4} = cos(\frac{3蟺}{2}) + isin(\frac{3蟺}{2})\),\n\n\(x_{5} = cos(\frac{11蟺}{6}) + isin(\frac{11蟺}{6})\)
04

Convert to rectangular form

Converting each solution from polar to rectangular form, we get the six solutions:\n\n\(x_{0} = \frac{\sqrt{3}}{2} + \frac{i}{2}\),\n\n\(x_{1} = i\),\n\n\(x_{2} = -\frac{\sqrt{3}}{2} + \frac{i}{2}\),\n\n\(x_{3} = -\frac{\sqrt{3}}{2} - \frac{i}{2}\),\n\n\(x_{4} = -i\),\n\n\(x_{5} = \frac{\sqrt{3}}{2} - \frac{i}{2}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Graph \(r_{1}\) and \(r_{2}\) in the same polar coordinate system. What is the relationship between the two graphs? $$r_{1}=2 \sin 3 \theta, r_{2}=2 \sin 3\left(\theta+\frac{\pi}{6}\right)$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I'm working with a polar equation that failed the symmetry test with respect to \(\theta=\frac{\pi}{2},\) so my graph will not have this kind of symmetry.

Use a graphing utility to graph each butterfly curve. Experiment with the range setting, particularly \(\theta\) step, to produce a butterfly of the best possible quality. $$\begin{aligned}&r=1.5^{\sin \theta}-2.5 \cos 4 \theta+\sin ^{7} \frac{\theta}{15} \quad \text { (Use } \quad \theta \min =0 \quad \text { and }\\\ &\theta \max =20 \pi .)\end{aligned}$$

Determine whether each statement makes sense or does not make sense, and explain your reasoning. There are no points on my graph of \(r^{2}=9 \cos 2 \theta\) for which \(\frac{\pi}{4}<\theta<\frac{3 \pi}{4}\)

Find \(\text {pro}_{\mathbf{w}} \mathbf{V}\) Then decompose v into two vectors, \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2},\) where \(\mathbf{v}_{1}\) is parallel to \(\mathbf{w}\) and \(\mathbf{v}_{2}\) is orthogonal to \(\mathbf{w}.\) $$\mathbf{v}=3 \mathbf{i}-2 \mathbf{j}, \quad \mathbf{w}=2 \mathbf{i}+\mathbf{j}$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.