If `S=underset(k=1) overset(10)sum(sin""(2pik)/11+icos""(2pik)/11)` then

Options

A. `S+barS=0`
B. `SbarS =1`
C. `sqrtS = +-1/sqrt2 (1+i)`
D. `S-barS =0`

Answer

To solve the problem, we need to evaluate the sum

\[ S = \sum_{k=1}^{10} \left( \sin\left(\frac{2\pi k}{11}\right) + i \cos\left(\frac{2\pi k}{11}\right) \right). \]

### Step 1: Rewrite the sum
We can factor out \(i\) from the sum:

\[ S = \sum_{k=1}^{10} \left( i \cos\left(\frac{2\pi k}{11}\right) + \sin\left(\frac{2\pi k}{11}\right) \right). \]

### Step 2: Use the identity for sine and cosine
Recall that \( \sin(x) + i \cos(x) = e^{ix} \) can be rewritten as:

\[ S = \sum_{k=1}^{10} \left( \cos\left(\frac{2\pi k}{11}\right) - i \sin\left(\frac{2\pi k}{11}\right) \right). \]

This can be expressed as:

\[ S = \sum_{k=1}^{10} e^{-i \frac{2\pi k}{11}}. \]

### Step 3: Recognize the geometric series
The sum \( \sum_{k=1}^{n} r^k \) for a geometric series can be calculated using the formula:

\[ S_n = a \frac{1 - r^n}{1 - r}, \]

where \(a\) is the first term and \(r\) is the common ratio. In our case, \(a = e^{-i \frac{2\pi}{11}}\) and \(r = e^{-i \frac{2\pi}{11}}\):

\[ S = e^{-i \frac{2\pi}{11}} \frac{1 - (e^{-i \frac{2\pi}{11}})^{10}}{1 - e^{-i \frac{2\pi}{11}}}. \]

### Step 4: Simplify the exponent
Calculating \( (e^{-i \frac{2\pi}{11}})^{10} \):

\[ (e^{-i \frac{2\pi}{11}})^{10} = e^{-i \frac{20\pi}{11}} = e^{-i \frac{20\pi}{11} + 2\pi} = e^{i \frac{2\pi}{11}}. \]

### Step 5: Substitute back into the sum
Now substituting back, we have:

\[ S = e^{-i \frac{2\pi}{11}} \frac{1 - e^{i \frac{2\pi}{11}}}{1 - e^{-i \frac{2\pi}{11}}}. \]

### Step 6: Simplify the expression
Using the identity \(1 - e^{ix} = -2i \sin\left(\frac{x}{2}\right)e^{i\frac{x}{2}}\):

\[ S = e^{-i \frac{2\pi}{11}} \frac{-2i \sin\left(\frac{\pi}{11}\right)e^{i\frac{\pi}{11}}}{-2i \sin\left(\frac{\pi}{11}\right)e^{-i\frac{\pi}{11}}} = e^{-i \frac{2\pi}{11}} e^{i \frac{\pi}{11}} = e^{-i \frac{\pi}{11}}. \]

### Step 7: Final result
Thus, we have:

\[ S = e^{-i \frac{\pi}{11}}. \]

### Step 8: Check the options
Now we check the options given in the problem.

1. \(S + S^* = 0\) is true since \(S^* = e^{i \frac{\pi}{11}}\).
2. \(S \cdot S^* = 1\) is also true since \(|S|^2 = 1\).
3. \(S - S^* = 0\) is false since \(S - S^* = -2i \sin\left(\frac{\pi}{11}\right) \neq 0\).

### Conclusion
The correct options are:
- Option A: True
- Option B: True
- Option C: False
- Option D: False