Find the value of `(1+ i)^(6) + (1-i)^(6)`

Options

A. `16i`
B. `0`
C. `-16i`
D. `1`

Explanation

To find the value of \((1 + i)^6 + (1 - i)^6\), we can follow these steps:

### Step 1: Express the complex numbers in polar form
The complex numbers \(1 + i\) and \(1 - i\) can be expressed in polar form.

For \(1 + i\):
- The modulus is \(|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}\).
- The argument (angle) is \(\tan^{-1}(1/1) = \frac{\pi}{4}\).

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

For \(1 - i\):
- The modulus is \(|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\).
- The argument is \(\tan^{-1}(-1/1) = -\frac{\pi}{4}\).

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

### Step 2: Raise to the power of 6
Using De Moivre's theorem, we can raise these complex numbers to the power of 6.

For \((1 + i)^6\):
\[ (1 + i)^6 = \left(\sqrt{2}\right)^6 \left( \cos\left(6 \cdot \frac{\pi}{4}\right) + i \sin\left(6 \cdot \frac{\pi}{4}\right) \right) \]
Calculating:
\[ \left(\sqrt{2}\right)^6 = 2^3 = 8 \]
And:
\[ 6 \cdot \frac{\pi}{4} = \frac{3\pi}{2} \]
Thus:
\[ (1 + i)^6 = 8 \left( \cos\left(\frac{3\pi}{2}\right) + i \sin\left(\frac{3\pi}{2}\right) \right) = 8(0 - i) = -8i \]

For \((1 - i)^6\):
\[ (1 - i)^6 = \left(\sqrt{2}\right)^6 \left( \cos\left(6 \cdot -\frac{\pi}{4}\right) + i \sin\left(6 \cdot -\frac{\pi}{4}\right) \right) \]
Calculating:
\[ (1 - i)^6 = 8 \left( \cos\left(-\frac{3\pi}{2}\right) + i \sin\left(-\frac{3\pi}{2}\right) \right) = 8(0 + i) = 8i \]

### Step 3: Add the results
Now, we add the two results:
\[ (1 + i)^6 + (1 - i)^6 = -8i + 8i = 0 \]

### Final Answer
Thus, the value of \((1 + i)^6 + (1 - i)^6\) is:
\[ \boxed{0} \]