13п 30√3(cos + i sin ¹3″) and 9 13TT Given the complex number 2₁ = 9 23 п 22 = 5( cos 2³+ i sin 23), express the result of 41 in rectangular form with fully 21 22 18 simplified fractions and radicals.
13п 30√3(cos + i sin ¹3″) and 9 13TT Given the complex number 2₁ = 9 23 п 22 = 5( cos 2³+ i sin 23), express the result of 41 in rectangular form with fully 21 22 18 simplified fractions and radicals.
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...
Related questions
Question
![### Problem Statement
Given the complex number \( z_1 = 30\sqrt{3} \left( \cos \frac{13\pi}{9} + i \sin \frac{13\pi}{9} \right) \) and
\( z_2 = 5 \left( \cos \frac{23\pi}{18} + i \sin \frac{23\pi}{18} \right) \), express the result of \( \frac{z_1}{z_2} \) in rectangular form with fully simplified fractions and radicals.
### Solution
To solve this problem, first note that complex numbers in polar form can be divided by subtracting their angles and dividing their magnitudes:
\[ \frac{z_1}{z_2} = \frac{30\sqrt{3}}{5} \left( \cos \left( \frac{13\pi}{9} - \frac{23\pi}{18} \right) + i \sin \left( \frac{13\pi}{9} - \frac{23\pi}{18} \right) \right) \]
1. Simplify the magnitude:
\[ \frac{30\sqrt{3}}{5} = 6\sqrt{3} \]
2. Simplify the angle:
\[ \frac{13\pi}{9} - \frac{23\pi}{18} = \frac{26\pi}{18} - \frac{23\pi}{18} = \frac{3\pi}{18} = \frac{\pi}{6} \]
3. Thus, we have:
\[ \frac{z_1}{z_2} = 6\sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \]
4. Using known values for cosine and sine:
\[ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \]
\[ \sin \frac{\pi}{6} = \frac{1}{2} \]
5. Substitute these values:
\[ \frac{z_1}{z_2} = 6\sqrt{3} \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \]
6.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e31dfb2-e24d-4f0b-ab76-e0eb03cfbad1%2Fc7d31f71-b0b7-40bb-a9f0-bc944cf4be13%2F7662oft_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Given the complex number \( z_1 = 30\sqrt{3} \left( \cos \frac{13\pi}{9} + i \sin \frac{13\pi}{9} \right) \) and
\( z_2 = 5 \left( \cos \frac{23\pi}{18} + i \sin \frac{23\pi}{18} \right) \), express the result of \( \frac{z_1}{z_2} \) in rectangular form with fully simplified fractions and radicals.
### Solution
To solve this problem, first note that complex numbers in polar form can be divided by subtracting their angles and dividing their magnitudes:
\[ \frac{z_1}{z_2} = \frac{30\sqrt{3}}{5} \left( \cos \left( \frac{13\pi}{9} - \frac{23\pi}{18} \right) + i \sin \left( \frac{13\pi}{9} - \frac{23\pi}{18} \right) \right) \]
1. Simplify the magnitude:
\[ \frac{30\sqrt{3}}{5} = 6\sqrt{3} \]
2. Simplify the angle:
\[ \frac{13\pi}{9} - \frac{23\pi}{18} = \frac{26\pi}{18} - \frac{23\pi}{18} = \frac{3\pi}{18} = \frac{\pi}{6} \]
3. Thus, we have:
\[ \frac{z_1}{z_2} = 6\sqrt{3} \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \]
4. Using known values for cosine and sine:
\[ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \]
\[ \sin \frac{\pi}{6} = \frac{1}{2} \]
5. Substitute these values:
\[ \frac{z_1}{z_2} = 6\sqrt{3} \left( \frac{\sqrt{3}}{2} + i \frac{1}{2} \right) \]
6.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 4 steps with 4 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning