If `i=sqrt(-1)`, then `(i^(n)+i^(-n), n in Z)` is equal to

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to evaluate \( i^n + i^{-n} \) for \( n \in \mathbb{Z} \), where \( i = \sqrt{-1} \). We will consider different cases based on the value of \( n \) modulo 4.

### Step-by-Step Solution:

1. **Understanding Powers of \( i \)**:
The powers of \( i \) cycle every four terms:
- \( i^0 = 1 \)
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \) (and so on)

2. **Case 1: \( n = 4m \)** (where \( m \) is an integer)
- Here, \( i^n = i^{4m} = (i^4)^m = 1^m = 1 \)
- Similarly, \( i^{-n} = i^{-4m} = (i^{-4})^m = 1^m = 1 \)
- Therefore, \( i^n + i^{-n} = 1 + 1 = 2 \)

3. **Case 2: \( n = 4m + 1 \)**
- Here, \( i^n = i^{4m+1} = i^{4m} \cdot i^1 = 1 \cdot i = i \)
- And \( i^{-n} = i^{-(4m+1)} = i^{-4m} \cdot i^{-1} = 1 \cdot (-i) = -i \)
- Therefore, \( i^n + i^{-n} = i - i = 0 \)

4. **Case 3: \( n = 4m + 2 \)**
- Here, \( i^n = i^{4m+2} = i^{4m} \cdot i^2 = 1 \cdot (-1) = -1 \)
- And \( i^{-n} = i^{-(4m+2)} = i^{-4m} \cdot i^{-2} = 1 \cdot (-1) = -1 \)
- Therefore, \( i^n + i^{-n} = -1 + (-1) = -2 \)

5. **Case 4: \( n = 4m + 3 \)**
- Here, \( i^n = i^{4m+3} = i^{4m} \cdot i^3 = 1 \cdot (-i) = -i \)
- And \( i^{-n} = i^{-(4m+3)} = i^{-4m} \cdot i^{-3} = 1 \cdot i = i \)
- Therefore, \( i^n + i^{-n} = -i + i = 0 \)

### Summary of Results:
- If \( n \equiv 0 \mod 4 \), then \( i^n + i^{-n} = 2 \)
- If \( n \equiv 1 \mod 4 \), then \( i^n + i^{-n} = 0 \)
- If \( n \equiv 2 \mod 4 \), then \( i^n + i^{-n} = -2 \)
- If \( n \equiv 3 \mod 4 \), then \( i^n + i^{-n} = 0 \)

### Final Answer:
The expression \( i^n + i^{-n} \) can take the values \( 2, 0, -2 \) depending on the value of \( n \mod 4 \).