Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Related questions
Question
![**Problem Statement:**
Divide the complex numbers. No \(i^2\) or spaces in answers please.
\[
\frac{(2-3i)}{(3+4i)} = \boxed{}
\]
**Explanation:**
The problem involves dividing two complex numbers. To solve, we multiply the numerator and the denominator by the conjugate of the denominator. For the denominator \(3+4i\), the conjugate is \(3-4i\).
**Steps to Solve:**
1. **Multiply the Numerator and Denominator by the Conjugate:**
\[
\frac{(2-3i)}{(3+4i)} \times \frac{(3-4i)}{(3-4i)}
\]
2. **Expand the Numerator:**
\[
(2-3i)(3-4i) = 2 \cdot 3 + 2 \cdot (-4i) + (-3i) \cdot 3 + (-3i) \cdot (-4i)
\]
Simplify each term:
\[
6 - 8i - 9i + 12i^2 \quad \text{(Note: \(i^2 = -1\))}
\]
\[
6 - 17i + 12(-1) = 6 - 17i - 12 = -6 - 17i
\]
3. **Expand the Denominator:**
\[
(3+4i)(3-4i) = 3 \cdot 3 + 3 \cdot (-4i) + 4i \cdot 3 + 4i \cdot (-4i)
\]
Simplify each term:
\[
9 - 12i + 12i - 16i^2
\]
\[
9 + 16 = 25
\]
4. **Combine Results:**
\[
\frac{-6 - 17i}{25}
\]
5. **Final Result:**
\[
-\frac{6}{25} - \frac{17}{25}i
\]
The division of the complex numbers yields:
\[
-\frac{6}{25} - \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F532d2df1-6bef-42be-bbaa-e1703c54e78f%2F154eb8f3-f69a-4923-bf00-354c9f2021d0%2F2870vhs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Divide the complex numbers. No \(i^2\) or spaces in answers please.
\[
\frac{(2-3i)}{(3+4i)} = \boxed{}
\]
**Explanation:**
The problem involves dividing two complex numbers. To solve, we multiply the numerator and the denominator by the conjugate of the denominator. For the denominator \(3+4i\), the conjugate is \(3-4i\).
**Steps to Solve:**
1. **Multiply the Numerator and Denominator by the Conjugate:**
\[
\frac{(2-3i)}{(3+4i)} \times \frac{(3-4i)}{(3-4i)}
\]
2. **Expand the Numerator:**
\[
(2-3i)(3-4i) = 2 \cdot 3 + 2 \cdot (-4i) + (-3i) \cdot 3 + (-3i) \cdot (-4i)
\]
Simplify each term:
\[
6 - 8i - 9i + 12i^2 \quad \text{(Note: \(i^2 = -1\))}
\]
\[
6 - 17i + 12(-1) = 6 - 17i - 12 = -6 - 17i
\]
3. **Expand the Denominator:**
\[
(3+4i)(3-4i) = 3 \cdot 3 + 3 \cdot (-4i) + 4i \cdot 3 + 4i \cdot (-4i)
\]
Simplify each term:
\[
9 - 12i + 12i - 16i^2
\]
\[
9 + 16 = 25
\]
4. **Combine Results:**
\[
\frac{-6 - 17i}{25}
\]
5. **Final Result:**
\[
-\frac{6}{25} - \frac{17}{25}i
\]
The division of the complex numbers yields:
\[
-\frac{6}{25} - \frac
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