The value of ` sum _(k = 0) ^(10) ( sin "" ( 2 k pi)/(11) + i cos "" (2 k pi)/( 11))` is

Options

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

Answer

To solve the problem, we need to evaluate the sum:

\[ S = \sum_{k=0}^{10} \left( \sin\left(\frac{2k\pi}{11}\right) + i \cos\left(\frac{2k\pi}{11}\right) \right) \]

### Step 1: Rewrite the terms using Euler's formula
Using Euler's formula, we can express the sine and cosine functions in terms of the exponential function:

\[ \sin\left(\frac{2k\pi}{11}\right) = \frac{e^{i\frac{2k\pi}{11}} - e^{-i\frac{2k\pi}{11}}}{2i} \]
\[ \cos\left(\frac{2k\pi}{11}\right) = \frac{e^{i\frac{2k\pi}{11}} + e^{-i\frac{2k\pi}{11}}}{2} \]

Thus, we can rewrite the sum \( S \):

\[ S = \sum_{k=0}^{10} \left( \frac{e^{i\frac{2k\pi}{11}} - e^{-i\frac{2k\pi}{11}}}{2i} + i \cdot \frac{e^{i\frac{2k\pi}{11}} + e^{-i\frac{2k\pi}{11}}}{2} \right) \]

### Step 2: Combine the terms
Combining the terms inside the sum gives us:

\[ S = \sum_{k=0}^{10} \left( \frac{e^{i\frac{2k\pi}{11}} - e^{-i\frac{2k\pi}{11}} + 2i^2 e^{i\frac{2k\pi}{11}} + 2i^2 e^{-i\frac{2k\pi}{11}}}{2i} \right) \]

Since \( i^2 = -1 \), we have:

\[ S = \sum_{k=0}^{10} \left( \frac{(1 - 2)e^{i\frac{2k\pi}{11}} + (1 + 2)e^{-i\frac{2k\pi}{11}}}{2i} \right) \]

This simplifies to:

\[ S = \sum_{k=0}^{10} \left( \frac{-e^{i\frac{2k\pi}{11}} + 3e^{-i\frac{2k\pi}{11}}}{2i} \right) \]

### Step 3: Factor out the common terms
We can factor out the common terms from the sum:

\[ S = \frac{1}{2i} \left( -\sum_{k=0}^{10} e^{i\frac{2k\pi}{11}} + 3\sum_{k=0}^{10} e^{-i\frac{2k\pi}{11}} \right) \]

### Step 4: Evaluate the sums
The sums \( \sum_{k=0}^{10} e^{i\frac{2k\pi}{11}} \) and \( \sum_{k=0}^{10} e^{-i\frac{2k\pi}{11}} \) are geometric series. The sum of a geometric series can be calculated using the formula:

\[ \text{Sum} = a \frac{1 - r^n}{1 - r} \]

where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

For \( \sum_{k=0}^{10} e^{i\frac{2k\pi}{11}} \):

- First term \( a = 1 \)
- Common ratio \( r = e^{i\frac{2\pi}{11}} \)
- Number of terms \( n = 11 \)

Thus:

\[ \sum_{k=0}^{10} e^{i\frac{2k\pi}{11}} = \frac{1 - (e^{i\frac{2\pi}{11}})^{11}}{1 - e^{i\frac{2\pi}{11}}} = \frac{1 - e^{2\pi i}}{1 - e^{i\frac{2\pi}{11}}} = \frac{0}{1 - e^{i\frac{2\pi}{11}}} = 0 \]

Similarly, for \( \sum_{k=0}^{10} e^{-i\frac{2k\pi}{11}} \):

\[ \sum_{k=0}^{10} e^{-i\frac{2k\pi}{11}} = 0 \]

### Step 5: Conclude the sum
Substituting these results back into our expression for \( S \):

\[ S = \frac{1}{2i} \left( -0 + 3 \cdot 0 \right) = 0 \]

Thus, the value of the sum is:

\[ \boxed{0} \]