Find the values of the following : (i) `i^(7)+i^(17)+i^(12)` (ii) `i^(11)+i^(-11)` (iii) `i^(3)+(1)/(i^(3))` (iv) `1+i^(2)+i^(6)+i^(8)`

Explanation

To solve the given problems involving powers of the imaginary unit \( i \), we first need to recall the fundamental properties of \( i \):

- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)

These powers repeat every four terms, so we can reduce any exponent modulo 4.

Now, let's solve each part step by step.

### (i) \( i^{7} + i^{17} + i^{12} \)

1. **Calculate \( i^{7} \)**:
\[ 7 \mod 4 = 3 \quad \Rightarrow \quad i^{7} = i^{3} = -i \]

2. **Calculate \( i^{17} \)**:
\[ 17 \mod 4 = 1 \quad \Rightarrow \quad i^{17} = i^{1} = i \]

3. **Calculate \( i^{12} \)**:
\[ 12 \mod 4 = 0 \quad \Rightarrow \quad i^{12} = i^{0} = 1 \]

4. **Combine the results**:
\[ i^{7} + i^{17} + i^{12} = -i + i + 1 = 1 \]

### (ii) \( i^{11} + i^{-11} \)

1. **Calculate \( i^{11} \)**:
\[ 11 \mod 4 = 3 \quad \Rightarrow \quad i^{11} = i^{3} = -i \]

2. **Calculate \( i^{-11} \)**:
\[ -11 \mod 4 = 1 \quad \Rightarrow \quad i^{-11} = i^{1} = i \]

3. **Combine the results**:
\[ i^{11} + i^{-11} = -i + i = 0 \]

### (iii) \( i^{3} + \frac{1}{i^{3}} \)

1. **Calculate \( i^{3} \)**:
\[ i^{3} = -i \]

2. **Calculate \( \frac{1}{i^{3}} \)**:
\[ \frac{1}{i^{3}} = \frac{1}{-i} = \frac{-i}{-i \cdot i} = \frac{-i}{1} = -i \]

3. **Combine the results**:
\[ i^{3} + \frac{1}{i^{3}} = -i - i = -2i \]

### (iv) \( 1 + i^{2} + i^{6} + i^{8} \)

1. **Calculate \( i^{2} \)**:
\[ i^{2} = -1 \]

2. **Calculate \( i^{6} \)**:
\[ 6 \mod 4 = 2 \quad \Rightarrow \quad i^{6} = i^{2} = -1 \]

3. **Calculate \( i^{8} \)**:
\[ 8 \mod 4 = 0 \quad \Rightarrow \quad i^{8} = i^{0} = 1 \]

4. **Combine the results**:
\[ 1 + i^{2} + i^{6} + i^{8} = 1 - 1 - 1 + 1 = 0 \]

### Final Answers:
(i) \( 1 \)
(ii) \( 0 \)
(iii) \( -2i \)
(iv) \( 0 \)