Evaluate: `" "(i)" "(i^(41)+(1)/(i^(71)))" "(ii)" "(i^(53)+(1)/(i^(53)))`

Answer

To solve the given problems, we will evaluate the expressions step by step.

### Part (i): Evaluate \( i^{41} + \frac{1}{i^{71}} \)

1. **Finding \( i^{41} \)**:
- The powers of \( i \) cycle every 4:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- To find \( i^{41} \), we calculate \( 41 \mod 4 \):
\[ 41 \div 4 = 10 \quad \text{(remainder 1)} \]
Thus, \( 41 \mod 4 = 1 \) and \( i^{41} = i^1 = i \).

2. **Finding \( i^{71} \)**:
- Similarly, calculate \( 71 \mod 4 \):
\[ 71 \div 4 = 17 \quad \text{(remainder 3)} \]
Thus, \( 71 \mod 4 = 3 \) and \( i^{71} = i^3 = -i \).

3. **Calculating \( \frac{1}{i^{71}} \)**:
- Since \( i^{71} = -i \):
\[ \frac{1}{i^{71}} = \frac{1}{-i} = -\frac{1}{i} \]
- To simplify \( -\frac{1}{i} \), we multiply the numerator and denominator by \( i \):
\[ -\frac{1}{i} \cdot \frac{i}{i} = -\frac{i}{i^2} = -\frac{i}{-1} = i \]

4. **Combining the results**:
- Now we have:
\[ i^{41} + \frac{1}{i^{71}} = i + i = 2i \]

### Part (ii): Evaluate \( i^{53} + \frac{1}{i^{53}} \)

1. **Finding \( i^{53} \)**:
- Calculate \( 53 \mod 4 \):
\[ 53 \div 4 = 13 \quad \text{(remainder 1)} \]
Thus, \( 53 \mod 4 = 1 \) and \( i^{53} = i^1 = i \).

2. **Calculating \( \frac{1}{i^{53}} \)**:
- Since \( i^{53} = i \):
\[ \frac{1}{i^{53}} = \frac{1}{i} \]
- To simplify \( \frac{1}{i} \), we multiply the numerator and denominator by \( i \):
\[ \frac{1}{i} \cdot \frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i \]

3. **Combining the results**:
- Now we have:
\[ i^{53} + \frac{1}{i^{53}} = i - i = 0 \]

### Final Answers:
- (i) \( i^{41} + \frac{1}{i^{71}} = 2i \)
- (ii) \( i^{53} + \frac{1}{i^{53}} = 0 \)