Question

In Problems 1-26, write the given number in the form a + ib. 1. 2i^3 - 3i^2 + 5i 2. 3i^5 - i^4 + 7i^3 - 10i^2 - 9 3. i^8 4. i^11 5. (5 - 9i) + (2 - 4i) 6. 3(4 - i) - 3(5 + 2i) 7. i(5 + 7i) 8. i(4 - i) + 4i(1 + 2i) 9. (2 - 3i)(4 + i) 10. (1/2 - 1/4i)(2/3 + 5/3i) 11. (2 + 3i)^2 12. (1 - i)^3

          In Problems 1-26, write the given number in the form a + ib.
1. 2i^3 - 3i^2 + 5i
2. 3i^5 - i^4 + 7i^3 - 10i^2 - 9
3. i^8
4. i^11
5. (5 - 9i) + (2 - 4i)
6. 3(4 - i) - 3(5 + 2i)
7. i(5 + 7i)
8. i(4 - i) + 4i(1 + 2i)
9. (2 - 3i)(4 + i)
10. (1/2 - 1/4i)(2/3 + 5/3i)
11. (2 + 3i)^2
12. (1 - i)^3

In Problems 1-26, write the given number in the form a + ib.
1. 2i^3 - 3i^2 + 5i
2. 3i^5 - i^4 + 7i^3 - 10i^2 - 9
3. i^8
4. i^11
5. (5 - 9i) + (2 - 4i)
6. 3(4 - i) - 3(5 + 2i)
7. i(5 + 7i)
8. i(4 - i) + 4i(1 + 2i)
9. (2 - 3i)(4 + i)
10. (1/2 - 1/4i)(2/3 + 5/3i)
11. (2 + 3i)^2
12. (1 - i)^3

Added by Catherine F.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

07/15/2023

Step 1

Recall that \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and these powers repeat cyclically. ### Problem 1: \(2i^3 - 3i^2 + 5i\) 1. Calculate \(i^3 = -i\). 2. Substitute: \(2(-i) - 3(-1) + 5i\). 3. Simplify: \(-2i + 3 + 5i\). 4. Combine like terms: \(3 + 3i\). ### Show more…

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In Problems 1-26, write the given number in the form a + ib. 1. 2i^3 - 3i^2 + 5i 2. 3i^5 - i^4 + 7i^3 - 10i^2 - 9 3. i^8 4. i^11 5. (5 - 9i) + (2 - 4i) 6. 3(4 - i) - 3(5 + 2i) 7. i(5 + 7i) 8. i(4 - i) + 4i(1 + 2i) 9. (2 - 3i)(4 + i) 10. (1/2 - 1/4i)(2/3 + 5/3i) 11. (2 + 3i)^2 12. (1 - i)^3