If n is an odd integer, then ` (1 + i)^(6n) + (1-i)^(6n)` is equal to

Options

A. 0
B. 2
C. `-2`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \( (1 + i)^{6n} + (1 - i)^{6n} \) where \( n \) is an odd integer.

### Step-by-Step Solution:

1. **Rewrite the expression**:
\[ (1 + i)^{6n} + (1 - i)^{6n} \]

2. **Convert to polar form**:
The complex numbers \( 1 + i \) and \( 1 - i \) can be expressed in polar form. The modulus and argument of \( 1 + i \) are:
- Modulus: \( |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2} \)
- Argument: \( \arg(1 + i) = \tan^{-1}(1) = \frac{\pi}{4} \)

Thus,
\[ 1 + i = \sqrt{2} \left( \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} \right) \]

Similarly for \( 1 - i \):
- Modulus: \( |1 - i| = \sqrt{2} \)
- Argument: \( \arg(1 - i) = \tan^{-1}(-1) = -\frac{\pi}{4} \)

Thus,
\[ 1 - i = \sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right) \right) \]

3. **Raise to the power \( 6n \)**:
Using De Moivre's theorem:
\[ (1 + i)^{6n} = (\sqrt{2})^{6n} \left( \cos\left(6n \cdot \frac{\pi}{4}\right) + i\sin\left(6n \cdot \frac{\pi}{4}\right) \right) \]
\[ (1 - i)^{6n} = (\sqrt{2})^{6n} \left( \cos\left(6n \cdot \left(-\frac{\pi}{4}\right)\right) + i\sin\left(6n \cdot \left(-\frac{\pi}{4}\right)\right) \right) \]

4. **Combine the two expressions**:
Since \( (\sqrt{2})^{6n} = 2^{3n} \), we can write:
\[ (1 + i)^{6n} + (1 - i)^{6n} = 2^{3n} \left( \cos\left(6n \cdot \frac{\pi}{4}\right) + i\sin\left(6n \cdot \frac{\pi}{4}\right) + \cos\left(-6n \cdot \frac{\pi}{4}\right) + i\sin\left(-6n \cdot \frac{\pi}{4}\right) \right) \]

The sine terms cancel out:
\[ = 2^{3n} \left( 2\cos\left(6n \cdot \frac{\pi}{4}\right) \right) \]

5. **Evaluate the cosine term**:
Since \( n \) is odd, \( 6n \) is also odd, and:
\[ 6n \cdot \frac{\pi}{4} = \frac{3n\pi}{2} \]
The cosine of \( \frac{3n\pi}{2} \) is \( 0 \) for odd \( n \).

6. **Final result**:
Thus,
\[ (1 + i)^{6n} + (1 - i)^{6n} = 2^{3n} \cdot 2 \cdot 0 = 0 \]

### Conclusion:
The final result is:
\[ (1 + i)^{6n} + (1 - i)^{6n} = 0 \]