/*! 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 13 Plot each complex number. Then w... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 13

Plot each complex number. Then write the complex number in polar form. You may express the argument in degrees or radians. $$-1-i$$

Short Answer

Expert verified
The complex number \(-1 - i\) can be represented in polar form as \( \sqrt{2} (\cos(225掳) + i \sin(225掳)) \) in degrees or \( \sqrt{2} (\cos(\frac{5蟺}{4}) + i \sin(\frac{5蟺}{4})) \) in radians.

Step by step solution

01

Plotting the Complex Number

Complex numbers can be plotted on a plane where the horizontal axis represents the real part and the vertical axis represents the imaginary part. For the given complex number \(-1 - i\), the point (-1,-1) can be plotted in the complex plane since the real part is -1 (a = -1) and the imaginary part is -1 (b = -1).
02

Calculating the magnitude

The magnitude (r) of the complex number can be calculated using the formula \( r = \sqrt {a^2 + b^2} \). So, substituting a = -1 and b = -1, we get \( r = \sqrt {(-1)^2 + (-1)^2} = \sqrt {2} \).
03

Computing the Argument

The argument (胃) of a complex number is the counterclockwise angle from the positive real axis to the line segment that joins the origin to the point. It is calculated using the formula \( 胃 = \tan^{-1} {b/a} \). Substituting b = -1 and a = -1, we find \( 胃 = \tan^{-1} {(-1)/(-1)} = 45掳 \) or \( \frac{蟺}{4} \) radians. Since the point is in the third quadrant, we add 180掳 or \( 蟺 \) to the angle. Hence, \( 胃 = 225掳 \) or \( \frac{5蟺}{4} \) radians.
04

Writing the Polar Form

The polar form of a complex number is given by \( r(\cos(胃) + i \sin(胃)) \). Substituting the values for r and 胃 we obtained earlier, the polar form of \(-1 - i\) is \( \sqrt{2} (\cos(225掳) + i \sin(225掳)) \) or \( \sqrt{2} (\cos(\frac{5蟺}{4}) + i \sin(\frac{5蟺}{4})) \).

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

Find the angle, in degrees, between \(\mathbf{v}\) and \(\mathbf{w}.\) $$\mathbf{v}=3 \cos \frac{5 \pi}{3} \mathbf{i}+3 \sin \frac{5 \pi}{3} \mathbf{j}, \quad \mathbf{w}=2 \cos \pi \mathbf{i}+2 \sin \pi \mathbf{j}$$

Let $$\mathbf{u}=-\mathbf{i}+\mathbf{j}, \quad \mathbf{v}=3 \mathbf{i}-2 \mathbf{j}, \quad \text { and } \quad \mathbf{w}=-5 \mathbf{j}$$ Find each specified scalar or vector. $$\operatorname{proj}_{\mathbf{u}}(\mathbf{v}+\mathbf{w})$$

Use the dot product to determine whether v and w are orthogonal. $$\mathbf{v}=\mathbf{i}+\mathbf{j}, \quad \mathbf{w}=\mathbf{i}-\mathbf{j}$$

I'm working with a unit vector, so its dot product with itself must be 1

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 \(\mathbf{w}.\) $$\mathbf{v}=\mathbf{i}+2 \mathbf{j}, \quad \mathbf{w}=3 \mathbf{i}+6 \mathbf{j}$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Geometry

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.