If `z_r` = `sinfrac(2pir)(11)-icosfrac(2rpi)(11)` then : the value of `sum_(r=0)^10 z_r` is equal to

Options

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

Answer

To solve the problem, we need to evaluate the sum \( \sum_{r=0}^{10} z_r \) where \( z_r = \sin\left(\frac{2\pi r}{11}\right) - i \cos\left(\frac{2\pi r}{11}\right) \).

### Step-by-step Solution:

1. **Rewrite \( z_r \)**:
\[ z_r = \sin\left(\frac{2\pi r}{11}\right) - i \cos\left(\frac{2\pi r}{11}\right) \]
We can factor out \(-i\):
\[ z_r = -i\left(\cos\left(\frac{2\pi r}{11}\right) - i \sin\left(\frac{2\pi r}{11}\right)\right) \]

2. **Recognize the expression as a complex exponential**:
The expression inside the parentheses can be rewritten using Euler's formula:
\[ z_r = -i e^{-i\frac{2\pi r}{11}} \]

3. **Sum the values of \( z_r \)**:
Now, we need to compute:
\[ \sum_{r=0}^{10} z_r = -i \sum_{r=0}^{10} e^{-i\frac{2\pi r}{11}} \]

4. **Identify the sum of roots of unity**:
The sum \( \sum_{r=0}^{10} e^{-i\frac{2\pi r}{11}} \) represents the sum of the 11th roots of unity. The roots of unity are given by:
\[ e^{i\frac{2\pi k}{n}} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]
In this case, \( n = 11 \).

5. **Use the property of roots of unity**:
The sum of all \( n \)th roots of unity is always zero:
\[ \sum_{r=0}^{10} e^{i\frac{2\pi r}{11}} = 0 \]

6. **Conclude the sum**:
Therefore, we have:
\[ \sum_{r=0}^{10} z_r = -i \cdot 0 = 0 \]

### Final Answer:
\[ \sum_{r=0}^{10} z_r = 0 \]