Let `z_(k)=cos(2kpi)/10+isin(2kpi)/10,k=1,2,………..,9`. Then, `1/10{|1-z_(1)||1-z_(2)|……|1-z_(9)|}` equals

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to evaluate the expression \( \frac{1}{10} \left| 1 - z_1 \right| \left| 1 - z_2 \right| \cdots \left| 1 - z_9 \right| \) where \( z_k = \cos\left(\frac{2k\pi}{10}\right) + i \sin\left(\frac{2k\pi}{10}\right) \) for \( k = 1, 2, \ldots, 9 \).

### Step-by-step Solution:

1. **Identify the values of \( z_k \)**:
The values \( z_k \) are the 10th roots of unity, given by:
\[ z_k = e^{\frac{2\pi i k}{10}} = \cos\left(\frac{2k\pi}{10}\right) + i \sin\left(\frac{2k\pi}{10}\right) \]
for \( k = 1, 2, \ldots, 9 \).

2. **Recognize the polynomial**:
The 10th roots of unity satisfy the polynomial equation:
\[ z^{10} - 1 = 0 \]
This can be factored as:
\[ z^{10} - 1 = (z - 1)(z - z_1)(z - z_2) \cdots (z - z_9) \]

3. **Evaluate at \( z = 1 \)**:
We need to evaluate \( |1 - z_k| \) for \( k = 1, 2, \ldots, 9 \):
\[ |1 - z_k| = |1 - e^{\frac{2\pi i k}{10}}| \]

4. **Use the formula for the magnitude**:
The magnitude can be computed as:
\[ |1 - z_k| = |1 - (\cos\left(\frac{2k\pi}{10}\right) + i \sin\left(\frac{2k\pi}{10}\right))| = \sqrt{(1 - \cos\left(\frac{2k\pi}{10}\right))^2 + \sin^2\left(\frac{2k\pi}{10}\right)} \]
Simplifying this gives:
\[ |1 - z_k| = \sqrt{2(1 - \cos\left(\frac{2k\pi}{10}\right))} = 2 \sin\left(\frac{k\pi}{10}\right) \]

5. **Calculate the product**:
Therefore, we have:
\[ \prod_{k=1}^{9} |1 - z_k| = \prod_{k=1}^{9} 2 \sin\left(\frac{k\pi}{10}\right) = 2^9 \prod_{k=1}^{9} \sin\left(\frac{k\pi}{10}\right) \]

6. **Use the known product of sines**:
The product \( \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}} \) for \( n = 10 \):
\[ \prod_{k=1}^{9} \sin\left(\frac{k\pi}{10}\right) = \frac{10}{2^9} \]

7. **Substituting back**:
Thus,
\[ \prod_{k=1}^{9} |1 - z_k| = 2^9 \cdot \frac{10}{2^9} = 10 \]

8. **Final calculation**:
Now substituting back into the original expression:
\[ \frac{1}{10} \prod_{k=1}^{9} |1 - z_k| = \frac{1}{10} \cdot 10 = 1 \]

### Conclusion:
The final answer is:
\[ \boxed{1} \]