Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Conversion of Complex Numbers to Polar Form
In this lesson, we will learn how to convert complex numbers into their polar form. We will use the given complex numbers \( z_1 \) and \( z_2 \) and express them in polar form. Remember, we will express the angle \( \theta \) in radians and ensure that \( 0 \leq \theta < 2\pi \).
**Given Complex Numbers:**
\[ z_1 = 4\sqrt{3} - 4i \]
\[ z_2 = -1 + i \]
### Conversion Process
1. **Calculate the Magnitude (\( r \)):**
The magnitude of a complex number \( z = a + bi \) is given by:
\[
r = \sqrt{a^2 + b^2}
\]
2. **Calculate the Angle (\( \theta \)):**
The angle \( \theta \) (in radians) is determined using the arctangent function:
\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\]
This angle needs to be adjusted depending on the quadrant in which the complex number is located.
3. **Express in Polar Form:**
The polar form of the complex number is given by:
\[
z = r \left( \cos(\theta) + i \sin(\theta) \right)
\]
**Example Solution as shown in the image:**
For \( z_1 = 4\sqrt{3} - 4i \):
- **Magnitude Calculation:**
\[
r = \sqrt{(4\sqrt{3})^2 + (-4)^2} = \sqrt{48 + 16} = \sqrt{64} = 8
\]
- **Angle Calculation:**
\[
\theta = \tan^{-1}\left(\frac{-4}{4\sqrt{3}}\right) = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}
\]
However, since the complex number is in the fourth quadrant, we need to add \( 2\pi \) to the angle to get:
\[
\theta = 2\pi - \frac{\pi}{6} = \frac{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f29774d-39d7-4c90-90ba-0f0924899c12%2F142501d1-fd5d-4266-9909-d672f64333c1%2Fs0741a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Conversion of Complex Numbers to Polar Form
In this lesson, we will learn how to convert complex numbers into their polar form. We will use the given complex numbers \( z_1 \) and \( z_2 \) and express them in polar form. Remember, we will express the angle \( \theta \) in radians and ensure that \( 0 \leq \theta < 2\pi \).
**Given Complex Numbers:**
\[ z_1 = 4\sqrt{3} - 4i \]
\[ z_2 = -1 + i \]
### Conversion Process
1. **Calculate the Magnitude (\( r \)):**
The magnitude of a complex number \( z = a + bi \) is given by:
\[
r = \sqrt{a^2 + b^2}
\]
2. **Calculate the Angle (\( \theta \)):**
The angle \( \theta \) (in radians) is determined using the arctangent function:
\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\]
This angle needs to be adjusted depending on the quadrant in which the complex number is located.
3. **Express in Polar Form:**
The polar form of the complex number is given by:
\[
z = r \left( \cos(\theta) + i \sin(\theta) \right)
\]
**Example Solution as shown in the image:**
For \( z_1 = 4\sqrt{3} - 4i \):
- **Magnitude Calculation:**
\[
r = \sqrt{(4\sqrt{3})^2 + (-4)^2} = \sqrt{48 + 16} = \sqrt{64} = 8
\]
- **Angle Calculation:**
\[
\theta = \tan^{-1}\left(\frac{-4}{4\sqrt{3}}\right) = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6}
\]
However, since the complex number is in the fourth quadrant, we need to add \( 2\pi \) to the angle to get:
\[
\theta = 2\pi - \frac{\pi}{6} = \frac{
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