If w is a complex cube root of unity, then the value of `w^(99)+w^(100)+w^(101)` is

Options

A. 1
B. `-1`
C. 3
D. 0

Answer

To solve the problem, we need to find the value of \( w^{99} + w^{100} + w^{101} \), where \( w \) is a complex cube root of unity. The complex cube roots of unity are \( 1, w, w^2 \), where \( w = e^{2\pi i / 3} \) and \( w^2 = e^{4\pi i / 3} \).

### Step-by-step Solution:

1. **Understand the properties of cube roots of unity**:
The cube roots of unity satisfy the equation \( w^3 = 1 \). This implies that:
\[ w^3 = 1 \quad \Rightarrow \quad w^0 = 1, \quad w^1 = w, \quad w^2 = w^2, \quad w^3 = 1, \quad w^4 = w, \quad w^5 = w^2, \quad w^6 = 1, \text{ and so on.} \]

2. **Reduce the exponents modulo 3**:
To simplify \( w^{99} + w^{100} + w^{101} \), we can reduce the exponents modulo 3:
- \( 99 \mod 3 = 0 \) so \( w^{99} = w^0 = 1 \)
- \( 100 \mod 3 = 1 \) so \( w^{100} = w^1 = w \)
- \( 101 \mod 3 = 2 \) so \( w^{101} = w^2 \)

3. **Substitute back into the expression**:
Now we can substitute these values back into the expression:
\[ w^{99} + w^{100} + w^{101} = 1 + w + w^2 \]

4. **Use the property of the sum of the cube roots of unity**:
From the properties of cube roots of unity, we know that:
\[ 1 + w + w^2 = 0 \]

5. **Final result**:
Therefore, we have:
\[ w^{99} + w^{100} + w^{101} = 0 \]

### Conclusion:
The value of \( w^{99} + w^{100} + w^{101} \) is \( \boxed{0} \).