If n is a multiple of 3, then ` 1 + omega ^(n) + omega ^(2n) =`

Options

A. 1
B. 2
C. 3
D. 0

Answer

To solve the problem, we need to evaluate the expression \( 1 + \omega^n + \omega^{2n} \) given that \( n \) is a multiple of 3.

### Step-by-Step Solution:

1. **Understanding \( \omega \)**:
- \( \omega \) is a primitive cube root of unity. The cube roots of unity are the solutions to the equation \( z^3 = 1 \). These roots are \( 1, \omega, \) and \( \omega^2 \), where \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{4\pi i / 3} \).
- The important property of these roots is that \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

2. **Using the property of \( n \)**:
- Since \( n \) is a multiple of 3, we can express \( n \) as \( n = 3k \) for some integer \( k \).

3. **Calculating \( \omega^n \) and \( \omega^{2n} \)**:
- Now, we can substitute \( n \) into the expression:
\[ \omega^n = \omega^{3k} = (\omega^3)^k = 1^k = 1 \]
- Similarly, for \( \omega^{2n} \):
\[ \omega^{2n} = \omega^{2(3k)} = \omega^{6k} = (\omega^3)^{2k} = 1^{2k} = 1 \]

4. **Substituting back into the expression**:
- Now substitute \( \omega^n \) and \( \omega^{2n} \) back into the original expression:
\[ 1 + \omega^n + \omega^{2n} = 1 + 1 + 1 = 3 \]

5. **Final Result**:
- Therefore, the value of \( 1 + \omega^n + \omega^{2n} \) when \( n \) is a multiple of 3 is \( 3 \).

### Conclusion:
The final answer is:
\[ 1 + \omega^n + \omega^{2n} = 3 \]