`{(1+cospi//8+isinipi//8)/(1+cospi//8-isinpi//8)}^(8)=`

Options

A. `1+i`
B. `1-i`
C. 1
D. `-1`

Explanation

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

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{1 + \cos \frac{\pi}{8} + i \sin \frac{\pi}{8}}{1 + \cos \frac{\pi}{8} - i \sin \frac{\pi}{8}} \]
We can use the identity \(1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}\) to rewrite \(1 + \cos \frac{\pi}{8}\):
\[ 1 + \cos \frac{\pi}{8} = 2 \cos^2 \frac{\pi}{16} \]
Thus, we can rewrite the expression as:
\[ \frac{2 \cos^2 \frac{\pi}{16} + i \sin \frac{\pi}{8}}{2 \cos^2 \frac{\pi}{16} - i \sin \frac{\pi}{8}} \]

### Step 2: Rewrite \(\sin \frac{\pi}{8}\)
Using the double angle formula for sine, we have:
\[ \sin \frac{\pi}{8} = 2 \sin \frac{\pi}{16} \cos \frac{\pi}{16} \]
Substituting this into our expression gives:
\[ \frac{2 \cos^2 \frac{\pi}{16} + i \cdot 2 \sin \frac{\pi}{16} \cos \frac{\pi}{16}}{2 \cos^2 \frac{\pi}{16} - i \cdot 2 \sin \frac{\pi}{16} \cos \frac{\pi}{16}} \]

### Step 3: Factor out common terms
We can factor out \(2 \cos \frac{\pi}{16}\) from both the numerator and the denominator:
\[ \frac{2 \cos \frac{\pi}{16} \left(\cos \frac{\pi}{16} + i \sin \frac{\pi}{16}\right)}{2 \cos \frac{\pi}{16} \left(\cos \frac{\pi}{16} - i \sin \frac{\pi}{16}\right)} \]
This simplifies to:
\[ \frac{\cos \frac{\pi}{16} + i \sin \frac{\pi}{16}}{\cos \frac{\pi}{16} - i \sin \frac{\pi}{16}} \]

### Step 4: Use Euler's formula
Using Euler's formula \(e^{i\theta} = \cos \theta + i \sin \theta\), we can rewrite the expression as:
\[ \frac{e^{i \frac{\pi}{16}}}{e^{-i \frac{\pi}{16}}} \]
This simplifies to:
\[ e^{i \frac{\pi}{16} + i \frac{\pi}{16}} = e^{i \frac{\pi}{8}} \]

### Step 5: Raise to the power of 8
Now we raise the expression to the power of 8:
\[ \left(e^{i \frac{\pi}{8}}\right)^{8} = e^{i \pi} \]

### Step 6: Evaluate \(e^{i \pi}\)
Using Euler's formula again, we know that:
\[ e^{i \pi} = \cos \pi + i \sin \pi = -1 + 0i = -1 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{-1} \]