`i^(2014)`=

Options

A. `i^(13)`
B. `i^(203)`
C. `i^(726)`
D. `i^(1993)`

Explanation

To solve the problem \( i^{2014} \), we first need to understand the powers of \( i \), where \( i \) is defined as the imaginary unit \( \sqrt{-1} \).

### Step-by-Step Solution:

1. **Understanding the Powers of \( i \)**:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = i^2 \cdot i = -1 \cdot i = -i \)
- \( i^4 = i^2 \cdot i^2 = -1 \cdot -1 = 1 \)

2. **Identifying the Pattern**:
- The powers of \( i \) repeat every four terms:
- \( i^1 = i \)
- \( i^2 = -1 \)
- \( i^3 = -i \)
- \( i^4 = 1 \)
- Thus, \( i^{n} \) can be determined by finding \( n \mod 4 \).

3. **Calculating \( 2014 \mod 4 \)**:
- To find \( i^{2014} \), we need to calculate \( 2014 \mod 4 \).
- Dividing \( 2014 \) by \( 4 \):
\[ 2014 \div 4 = 503 \quad \text{(with a remainder of 2)} \]
- Thus, \( 2014 \mod 4 = 2 \).

4. **Finding \( i^{2014} \)**:
- Since \( 2014 \mod 4 = 2 \), we have:
\[ i^{2014} = i^2 \]
- From our earlier calculations, we know that:
\[ i^2 = -1 \]

5. **Conclusion**:
- Therefore, \( i^{2014} = -1 \).

### Final Answer:
\[ i^{2014} = -1 \]