If `i= sqrt(-1)`, then which of the following has the greatest value ?

Options

A. `i^(4)+i^(3)+i^(2)+i`
B. `i^(8)+i^(6)+i^(4)+i^(2)`
C. `i^(12)+i^(9)+i^(6)+i^(3)`
D. `i^(16)+i^(12)+i^(8)+i^(4)`

Explanation

To solve the problem, we need to evaluate the expressions given in the options and determine which one has the greatest value. We know that \( i = \sqrt{-1} \) and we can derive the powers of \( i \) as follows:

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

From this, we see that the powers of \( i \) repeat every four terms:
- \( i^5 = i \)
- \( i^6 = -1 \)
- \( i^7 = -i \)
- \( i^8 = 1 \)
- And so on...

2. **Evaluate each option**:
- **Option A**: \( i^4 + i^3 + i^2 + i^1 \)
- \( = 1 + (-i) + (-1) + i \)
- \( = 1 - 1 + (-i + i) = 0 \)

- **Option B**: \( i^5 + i^6 + i^7 + i^8 \)
- \( = i + (-1) + (-i) + 1 \)
- \( = (i - i) + (-1 + 1) = 0 \)

- **Option C**: \( i^6 + i^7 + i^8 + i^9 \)
- \( = -1 + (-i) + 1 + i \)
- \( = (-1 + 1) + (-i + i) = 0 \)

- **Option D**: \( i^{16} + i^{12} + i^{8} + i^{4} \)
- \( = 1 + 1 + 1 + 1 \) (since \( i^4 = 1 \), \( i^8 = 1 \), \( i^{12} = 1 \), and \( i^{16} = 1 \))
- \( = 1 + 1 + 1 + 1 = 4 \)

3. **Compare the results**:
- Option A = 0
- Option B = 0
- Option C = 0
- Option D = 4

Thus, the option with the greatest value is **Option D**, which equals 4.