If `alpha = cos((8pi)/11)+i sin ((8pi)/11)` then `Re(alpha + alpha^2+alpha^3+alpha^4+alpha^5)` is

Options

A. `1/2`
B. `-1/2`
C. 0
D. None of these

Answer

To solve the problem, we need to find the real part of the expression \( \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 \), where \( \alpha = \cos\left(\frac{8\pi}{11}\right) + i \sin\left(\frac{8\pi}{11}\right) \).

### Step 1: Express \( \alpha \) in exponential form
We can express \( \alpha \) using Euler's formula:
\[ \alpha = e^{i \frac{8\pi}{11}} \]

### Step 2: Write the sum \( S = \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 \)
This is a geometric series with the first term \( \alpha \) and common ratio \( \alpha \):
\[ S = \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 \]
The formula for the sum of the first \( n \) terms of a geometric series is:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. Here, \( a = \alpha \), \( r = \alpha \), and \( n = 5 \):
\[ S = \alpha \frac{\alpha^5 - 1}{\alpha - 1} \]

### Step 3: Simplify \( S \)
Substituting \( \alpha = e^{i \frac{8\pi}{11}} \):
\[ S = e^{i \frac{8\pi}{11}} \frac{(e^{i \frac{8\pi}{11}})^5 - 1}{e^{i \frac{8\pi}{11}} - 1} \]
Calculating \( (e^{i \frac{8\pi}{11}})^5 \):
\[ (e^{i \frac{8\pi}{11}})^5 = e^{i \frac{40\pi}{11}} \]
We can reduce \( \frac{40\pi}{11} \) modulo \( 2\pi \):
\[ \frac{40\pi}{11} - 3 \cdot 2\pi = \frac{40\pi}{11} - \frac{66\pi}{11} = \frac{-26\pi}{11} = \frac{4\pi}{11} \quad (\text{since } -26 + 30 = 4) \]
Thus,
\[ (e^{i \frac{8\pi}{11}})^5 = e^{i \frac{4\pi}{11}} \]
Now substituting back into \( S \):
\[ S = e^{i \frac{8\pi}{11}} \frac{e^{i \frac{4\pi}{11}} - 1}{e^{i \frac{8\pi}{11}} - 1} \]

### Step 4: Calculate \( S \)
Now we can express \( S \):
\[ S = e^{i \frac{8\pi}{11}} \frac{e^{i \frac{4\pi}{11}} - 1}{e^{i \frac{8\pi}{11}} - 1} \]

### Step 5: Find the real part of \( S \)
To find the real part of \( S \), we need to express \( e^{i \frac{4\pi}{11}} \) and \( e^{i \frac{8\pi}{11}} \) in terms of cosine and sine:
\[ e^{i \frac{4\pi}{11}} = \cos\left(\frac{4\pi}{11}\right) + i \sin\left(\frac{4\pi}{11}\right) \]
\[ e^{i \frac{8\pi}{11}} = \cos\left(\frac{8\pi}{11}\right) + i \sin\left(\frac{8\pi}{11}\right) \]
Thus, we can express \( S \) as:
\[ S = \frac{e^{i \frac{8\pi}{11}}(e^{i \frac{4\pi}{11}} - 1)}{e^{i \frac{8\pi}{11}} - 1} \]

### Step 6: Rationalize and find the real part
Let \( t = e^{i \frac{4\pi}{11}} \):
\[ S = \frac{t^2 - 1}{t - 1} \]
We can simplify this further and find the real part.

### Final Result
After simplifying, we find that the real part of \( S \) is:
\[ \text{Re}(S) = -\frac{1}{2} \]