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
100%
![The problem presented is:
Express \(\left( \frac{\sqrt{5}}{5} - \frac{\sqrt{5}}{2}i \right)^2\) in simplest \(a + bi\) form.
To solve this, expand the expression using the formula for the square of a binomial:
\[
(a - bi)^2 = a^2 - 2abi + (bi)^2
\]
Substitute \(a = \frac{\sqrt{5}}{5}\) and \(b = \frac{\sqrt{5}}{2}\):
1. Calculate \(a^2\):
\[
a^2 = \left(\frac{\sqrt{5}}{5}\right)^2 = \frac{5}{25} = \frac{1}{5}
\]
2. Calculate \((bi)^2\):
\[
(bi)^2 = \left(\frac{\sqrt{5}}{2}i\right)^2 = -\left(\frac{5}{4}\right) = -\frac{5}{4}
\]
3. Calculate \(-2abi\):
\[
-2abi = -2 \times \frac{\sqrt{5}}{5} \times \frac{\sqrt{5}}{2}i = -\frac{2 \times 5}{10}i = -i
\]
Combine these results:
\[
\left(\frac{\sqrt{5}}{5} - \frac{\sqrt{5}}{2}i\right)^2 = \frac{1}{5} - i - \frac{5}{4}
\]
Simplify the real and imaginary parts:
- Real part: \(\frac{1}{5} - \frac{5}{4} = \frac{4}{20} - \frac{25}{20} = -\frac{21}{20}\)
- Imaginary part: \(-i\)
Thus, the expression in the simplest \(a + bi\) form is:
\[
-\frac{21}{20} - i
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8cc14db3-0b21-46fc-8a3d-1d0dbc6bee86%2F620cec17-3e8e-42df-a97f-828715c6c252%2Fha6x5b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The problem presented is:
Express \(\left( \frac{\sqrt{5}}{5} - \frac{\sqrt{5}}{2}i \right)^2\) in simplest \(a + bi\) form.
To solve this, expand the expression using the formula for the square of a binomial:
\[
(a - bi)^2 = a^2 - 2abi + (bi)^2
\]
Substitute \(a = \frac{\sqrt{5}}{5}\) and \(b = \frac{\sqrt{5}}{2}\):
1. Calculate \(a^2\):
\[
a^2 = \left(\frac{\sqrt{5}}{5}\right)^2 = \frac{5}{25} = \frac{1}{5}
\]
2. Calculate \((bi)^2\):
\[
(bi)^2 = \left(\frac{\sqrt{5}}{2}i\right)^2 = -\left(\frac{5}{4}\right) = -\frac{5}{4}
\]
3. Calculate \(-2abi\):
\[
-2abi = -2 \times \frac{\sqrt{5}}{5} \times \frac{\sqrt{5}}{2}i = -\frac{2 \times 5}{10}i = -i
\]
Combine these results:
\[
\left(\frac{\sqrt{5}}{5} - \frac{\sqrt{5}}{2}i\right)^2 = \frac{1}{5} - i - \frac{5}{4}
\]
Simplify the real and imaginary parts:
- Real part: \(\frac{1}{5} - \frac{5}{4} = \frac{4}{20} - \frac{25}{20} = -\frac{21}{20}\)
- Imaginary part: \(-i\)
Thus, the expression in the simplest \(a + bi\) form is:
\[
-\frac{21}{20} - i
\]
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 3 steps with 3 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