If `z_r` = `sinfrac(2pir)(11)-icosfrac(2rpi)(11)` then : the value of `sum_(r=1)^10 z_r` is equal to

Options

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

Answer

To solve the problem, we need to calculate the value of the sum \( \sum_{r=1}^{10} z_r \) 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**:
The expression inside the parentheses is the exponential form of complex numbers:
\[ z_r = -i e^{i\frac{2\pi r}{11}} \]

3. **Calculate the sum**:
Now we need to calculate:
\[ \sum_{r=1}^{10} z_r = -i \sum_{r=1}^{10} e^{i\frac{2\pi r}{11}} \]
The sum \( \sum_{r=0}^{10} e^{i\frac{2\pi r}{11}} \) represents the sum of the 11th roots of unity, which is known to be zero:
\[ \sum_{r=0}^{10} e^{i\frac{2\pi r}{11}} = 0 \]

4. **Separate the missing term**:
Since we are summing from \( r=1 \) to \( r=10 \), we can express the sum as:
\[ \sum_{r=1}^{10} e^{i\frac{2\pi r}{11}} = -e^{i\frac{2\pi \cdot 0}{11}} = -1 \]

5. **Final calculation**:
Therefore, substituting back into our sum:
\[ \sum_{r=1}^{10} z_r = -i(-1) = i \]

### Conclusion:
The value of \( \sum_{r=1}^{10} z_r \) is \( i \).