`((1 + i)/( sqrt(2)))^(8) + (( 1 - i)/( sqrt(2)))^(8) = `

Options

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

Answer

To solve the expression \(\left(\frac{1 + i}{\sqrt{2}}\right)^{8} + \left(\frac{1 - i}{\sqrt{2}}\right)^{8}\), we can follow these steps:

### Step 1: Convert to Polar Form
We start by converting the complex numbers \(1 + i\) and \(1 - i\) into polar form.

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

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

For \(1 - i\):
- The modulus is the same, \(|1 - i| = \sqrt{2}\).
- The argument is \(\tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}\).

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

### Step 2: Substitute into the Expression
Now we substitute these polar forms into the original expression:
\[ \left(\frac{1 + i}{\sqrt{2}}\right)^{8} = \left(\frac{\sqrt{2} e^{i\frac{\pi}{4}}}{\sqrt{2}}\right)^{8} = \left(e^{i\frac{\pi}{4}}\right)^{8} = e^{i 2\pi}. \]
\[ \left(\frac{1 - i}{\sqrt{2}}\right)^{8} = \left(\frac{\sqrt{2} e^{-i\frac{\pi}{4}}}{\sqrt{2}}\right)^{8} = \left(e^{-i\frac{\pi}{4}}\right)^{8} = e^{-i 2\pi}. \]

### Step 3: Combine the Results
Now we combine the results:
\[ e^{i 2\pi} + e^{-i 2\pi}. \]

### Step 4: Use Euler's Formula
Using Euler's formula, we know that:
\[ e^{i\theta} + e^{-i\theta} = 2\cos(\theta). \]
Thus:
\[ e^{i 2\pi} + e^{-i 2\pi} = 2\cos(2\pi). \]

### Step 5: Evaluate the Cosine
Since \(\cos(2\pi) = 1\):
\[ 2\cos(2\pi) = 2 \cdot 1 = 2. \]

### Final Result
Therefore, the value of the expression is:
\[ \boxed{2}. \]