If n is an integer, not a multiple of 3, the sum of `w^n+w^(2n),w` being a nonreal complex cube root of unity , is

Options

A. 2
B. -1
C. 2 or -1
D. none of these

Answer

To solve the problem, we need to find the sum \( \omega^n + \omega^{2n} \) where \( \omega \) is a non-real complex cube root of unity and \( n \) is an integer not a multiple of 3.

### Step-by-step Solution:

1. **Understanding Cube Roots of Unity**:
The cube roots of unity are the solutions to the equation \( x^3 = 1 \). They are:
\[ 1, \quad \omega, \quad \omega^2 \]
where \( \omega = e^{2\pi i / 3} = -\frac{1}{2} + i \frac{\sqrt{3}}{2} \) and \( \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - i \frac{\sqrt{3}}{2} \). The important properties are:
\[ 1 + \omega + \omega^2 = 0 \quad \text{and} \quad \omega^3 = 1 \]

2. **Considering the Cases for \( n \)**:
Since \( n \) is not a multiple of 3, it can be expressed in two forms:
- \( n = 3m + 1 \)
- \( n = 3m + 2 \)

3. **Case 1: \( n = 3m + 1 \)**:
\[ \omega^n = \omega^{3m + 1} = \omega^{3m} \cdot \omega = 1 \cdot \omega = \omega \]
\[ \omega^{2n} = \omega^{2(3m + 1)} = \omega^{6m + 2} = \omega^{6m} \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2 \]
Therefore, the sum becomes:
\[ \omega^n + \omega^{2n} = \omega + \omega^2 \]
Using the property \( 1 + \omega + \omega^2 = 0 \), we find:
\[ \omega + \omega^2 = -1 \]

4. **Case 2: \( n = 3m + 2 \)**:
\[ \omega^n = \omega^{3m + 2} = \omega^{3m} \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2 \]
\[ \omega^{2n} = \omega^{2(3m + 2)} = \omega^{6m + 4} = \omega^{6m} \cdot \omega^4 = 1 \cdot \omega^4 \]
Since \( \omega^4 = \omega \) (because \( \omega^3 = 1 \)), we have:
\[ \omega^{2n} = \omega \]
Thus, the sum becomes:
\[ \omega^n + \omega^{2n} = \omega^2 + \omega \]
Again, using the property \( 1 + \omega + \omega^2 = 0 \):
\[ \omega^2 + \omega = -1 \]

5. **Conclusion**:
In both cases, whether \( n = 3m + 1 \) or \( n = 3m + 2 \), we find that:
\[ \omega^n + \omega^{2n} = -1 \]

### Final Answer:
The sum \( \omega^n + \omega^{2n} \) is \( -1 \).