chap 4.7 imaginary numbers day 1
simplify the following.
1. $i^5 = $
2. $i^7 = $
3. $i^{12} = $
4. $i^{15} = $
5. $i^{19} = $
6. $i^{17} = $
7. $i^{23} = $
8. $i^{24} = $
9. $i^{53} = $
10. $i^{56} = $
11. $i^{79} = $
12. $i^{85} = $
13. $i^{66} = $
14. $i^{92} = $
15. $i^{94} = $
16. $i^{105} = $
17. $i^{59} = $
18. $i^{20} = $
19. $i^{11} = $

Question

Accepted Answer

To solve these problems, we use the properties of the imaginary unit $ i $, where $ i = \sqrt{-1} $, $ i^2 = -1 $, $ i^3 = -i $, $ i^4 = 1 $, and the powers of $ i $ repeat every 4. So we can simplify $ i^n $ by finding $ n \mod 4 $.

### 1. $ i^5 $
## Step 1: Find the remainder of 5 divided by 4
$ 5 \div 4 = 1 $ with a remainder of $ 1 $, so $ 5 \mod 4 = 1 $.
## Step 2: Simplify $ i^5 $
Since $ i^5 = i^{4 + 1} = i^4 \cdot i^1 $, and $ i^4 = 1 $, we have $ i^5 = 1 \cdot i = i $.
# Answer: $ i $

### 2. $ i^7 $
## Step 1: Find the remainder of 7 divided by 4
$ 7 \div 4 = 1 $ with a remainder of $ 3 $, so $ 7 \mod 4 = 3 $.
## Step 2: Simplify $ i^7 $
Since $ i^7 = i^{4 + 3} = i^4 \cdot i^3 $, and $ i^4 = 1 $, $ i^3 = -i $, we have $ i^7 = 1 \cdot (-i) = -i $.
# Answer: $ -i $

### 3. $ i^{12} $
## Step 1: Find the remainder of 12 divided by 4
$ 12 \div 4 = 3 $ with a remainder of $ 0 $, so $ 12 \mod 4 = 0 $. When the remainder is 0, we use $ i^4 $.
## Step 2: Simplify $ i^{12} $
Since $ i^{12} = (i^4)^3 $, and $ i^4 = 1 $, we have $ i^{12} = 1^3 = 1 $.
# Answer: $ 1 $

### 4. $ i^{15} $
## Step 1: Find the remainder of 15 divided by 4
$ 15 \div 4 = 3 $ with a remainder of $ 3 $, so $ 15 \mod 4 = 3 $.
## Step 2: Simplify $ i^{15} $
Since $ i^{15} = i^{4 \cdot 3 + 3} = (i^4)^3 \cdot i^3 $, and $ i^4 = 1 $, $ i^3 = -i $, we have $ i^{15} = 1^3 \cdot (-i) = -i $.
# Answer: $ -i $

### 5. $ i^{19} $
## Step 1: Find the remainder of 19 divided by 4
$ 19 \div 4 = 4 $ with a remainder of $ 3 $, so $ 19 \mod 4 = 3 $.
## Step 2: Simplify $ i^{19} $
Since $ i^{19} = i^{4 \cdot 4 + 3} = (i^4)^4 \cdot i^3 $, and $ i^4 = 1 $, $ i^3 = -i $, we have $ i^{19} = 1^4 \cdot (-i) = -i $.
# Answer: $ -i $

### 6. $ i^{17} $
## Step 1: Find the remainder of 17 divided by 4
$ 17 \div 4 = 4 $ with a remainder of $ 1 $, so $ 17 \mod 4 = 1 $.
## Step 2: Simplify $ i^{17} $
Since $ i^{17} = i^{4 \cdot 4 + 1} = (i^4)^4 \cdot i^1 $, and $ i^4 = 1 $, we have $ i^{17} = 1^4 \cdot i = i $.
# Answer: $ i $

### 7. $ i^{23} $
## Step 1: Find the remainder of 23 divided by 4
$ 23 \div 4 = 5 $ with a remainder of $ 3 $, so $ 23 \mod 4 = 3 $.
## Step 2: Simplify $ i^{23} $
Since $ i^{23} = i^{4 \cdot 5 + 3} = (i^4)^5 \cdot i^3 $, and $ i^4 = 1 $, $ i^3 = -i $, we have $ i^{23} = 1^5 \cdot (-i) = -i $.
# Answer: $ -i $

### 8. $ i^{24} $
## Step 1: Find the remainder of 24 divided by 4
$ 24 \div 4 = 6 $ with a remainder of $ 0 $, so $ 24 \mod 4 = 0 $.
## Step 2: Simplify $ i^{24} $
Since $ i^{24} = (i^4)^6 $, and $ i^4 = 1 $, we have $ i^{24} = 1^6 = 1 $.
# Answer: $ 1 $

### 9. $ i^{53} $
## Step 1: Find the remainder of 53 divided by 4
$ 53 \div 4 = 13 $ with a remainder of $ 1 $, so $ 53 \mod 4 = 1 $.
## Step 2: Simplify $ i^{53} $
Since $ i^{53} = i^{4 \cdot 13 + 1} = (i^4)^{13} \cdot i^1 $, and $ i^4 = 1 $, we have $ i^{53} = 1^{13} \cdot i = i $.
# Answer: $ i $

### 10. $ i^{56} $
## Step 1: Find the remainder of 56 divided by 4
$ 56 \div 4 = 14 $ with a remainder of $ 0 $, so $ 56 \mod 4 = 0 $.
## Step 2: Simplify $ i^{56} $
Since $ i^{56} = (i^4)^{14} $, and $ i^4 = 1 $, we have $ i^{56} = 1^{14} = 1 $.
# Answer: $ 1 $

### 11. $ i^{79} $
## Step 1: Find the remainder of 79 divided by 4
$ 79 \div 4 = 19 $ with a remainder of $ 3 $, so $ 79 \mod 4 = 3 $.
## Step 2: Simplify $ i^{79} $
Since $ i^{79} = i^{4 \cdot 19 + 3} = (i^4)^{19} \cdot i^3 $, and $ i^4 = 1 $, $ i^3 = -i $, we have $ i^{79} = 1^{19} \cdot (-i) = -i $.
# Answer: $ -i $

### 12. $ i^{85} $
## Step 1: Find the remainder of 85 divided by 4
$ 85 \div 4 = 21 $ with a remainder of $ 1 $, so $ 85 \mod 4 = 1 $.
## Step 2: Simplify $ i^{85} $
Since $ i^{85} = i^{4 \cdot 21 + 1} = (i^4)^{21} \cdot i^1 $, and $ i^4 = 1 $, we have $ i^{85} = 1^{21} \cdot i = i $.
# Answer: $ i $

### 13. $ i^{66} $
## Step 1: Find the remainder of 66 divided by 4
$ 66 \div 4 = 16 $ with a remainder of $ 2 $, so $ 66 \mod 4 = 2 $.
## Step 2: Simplify $ i^{66} $
Since $ i^{66} = i^{4 \cdot 16 + 2} = (i^4)^{16} \cdot i^2 $, and $ i^4 = 1 $, $ i^2 = -1 $, we have $ i^{66} = 1^{16} \cdot (-1) = -1 $.
# Answer: $ -1 $

### 14. $ i^{92} $
## Step 1: Find the remainder of 92 divided by 4
$ 92 \div 4 = 23 $ with a remainder of $ 0 $, so $ 92 \mod 4 = 0 $.
## Step 2: Simplify $ i^{92} $
Since $ i^{92} = (i^4)^{23} $, and $ i^4 = 1 $, we have $ i^{92} = 1^{23} = 1 $.
# Answer: $ 1 $

### 15. $ i^{94} $
## Step 1: Find the remainder of 94 divided by 4
$ 94 \div 4 = 23 $ with a remainder of $ 2 $, so $ 94 \mod 4 = 2 $.
## Step 2: Simplify $ i^{94} $
Since $ i^{94} = i^{4 \cdot 23 + 2} = (i^4)^{23} \cdot i^2 $, and $ i^4 = 1 $, $ i^2 = -1 $, we have $ i^{94} = 1^{23} \cdot (-1) = -1 $.
# Answer: $ -1 $

### 16. $ i^{105} $
## Step 1: Find the remainder of 105 divided by 4
$ 105 \div 4 = 26 $ with a remainder of $ 1 $, so $ 105 \mod 4 = 1 $.
## Step 2: Simplify $ i^{105} $
Since $ i^{105} = i^{4 \cdot 26 + 1} = (i^4)^{26} \cdot i^1 $, and $ i^4 = 1 $, we have $ i^{105} = 1^{26} \cdot i = i $.
# Answer: $ i $

### 17. $ i^{59} $
## Step 1: Find the remainder of 59 divided by 4
$ 59 \div 4 = 14 $ with a remainder of $ 3 $, so $ 59 \mod 4 = 3 $.
## Step 2: Simplify $ i^{59} $
Since $ i^{59} = i^{4 \cdot 14 + 3} = (i^4)^{14} \cdot i^3 $, and $ i^4 = 1 $, $ i^3 = -i $, we have $ i^{59} = 1^{14} \cdot (-i) = -i $.
# Answer: $ -i $

### 18. $ i^{20} $
## Step 1: Find the remainder of 20 divided by 4
$ 20 \div 4 = 5 $ with a remainder of $ 0 $, so $ 20 \mod 4 = 0 $.
## Step 2: Simplify $ i^{20} $
Since $ i^{20} = (i^4)^5 $, and $ i^4 = 1 $, we have $ i^{20} = 1^5 = 1 $.
# Answer: $ 1 $

### 19. $ i^{11} $
## Step 1: Find the remainder of 11 divided by 4
$ 11 \div 4 = 2 $ with a remainder of $ 3 $, so $ 11 \mod 4 = 3 $.
## Step 2: Simplify $ i^{11} $
Since $ i^{11} = i^{4 \cdot 2 + 3} = (i^4)^2 \cdot i^3 $, and $ i^4 = 1 $, $ i^3 = -i $, we have $ i^{11} = 1^2 \cdot (-i) = -i $.
# Answer: $ -i $

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QUESTION IMAGE

Question

Explanation:

1. \( i^5 \)

Step 1: Find the remainder of 5 divided by 4

Step 2: Simplify \( i^5 \)

Answer:

2. \( i^7 \)

Step 1: Find the remainder of 7 divided by 4

Step 2: Simplify \( i^7 \)