Let `z_(k) = cos ((2kpi)/(7))+i sin((2kpi)/(7)),"for k" = 1, 2, ..., 6`, then `log_(7)|1-z_(1)|+log_(7)|1-z_(2)|+....+ log_(7)|1-z_(6)| ` is equal to ________

Answer

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

\[ \log_{7}|1 - z_{1}| + \log_{7}|1 - z_{2}| + \ldots + \log_{7}|1 - z_{6}| \]

where

\[ z_{k} = \cos\left(\frac{2k\pi}{7}\right) + i \sin\left(\frac{2k\pi}{7}\right) \quad \text{for } k = 1, 2, \ldots, 6. \]

### Step 1: Identify the values of \( z_k \)

The values of \( z_k \) are the 7th roots of unity (excluding \( z_0 = 1 \)). Specifically, we have:

- \( z_1 = e^{i \frac{2\pi}{7}} \)
- \( z_2 = e^{i \frac{4\pi}{7}} \)
- \( z_3 = e^{i \frac{6\pi}{7}} \)
- \( z_4 = e^{i \frac{8\pi}{7}} \)
- \( z_5 = e^{i \frac{10\pi}{7}} \)
- \( z_6 = e^{i \frac{12\pi}{7}} \)

### Step 2: Calculate \( |1 - z_k| \)

The magnitude \( |1 - z_k| \) can be calculated using the formula:

\[ |1 - z_k| = |1 - e^{i \frac{2k\pi}{7}}| = \sqrt{(1 - \cos\left(\frac{2k\pi}{7}\right))^2 + \sin^2\left(\frac{2k\pi}{7}\right)} \]

Using the identity \( \sin^2 x + \cos^2 x = 1 \):

\[ |1 - z_k| = \sqrt{(1 - \cos\left(\frac{2k\pi}{7}\right))^2 + (1 - \cos^2\left(\frac{2k\pi}{7}\right))} \]

This simplifies to:

\[ |1 - z_k| = \sqrt{2 - 2\cos\left(\frac{2k\pi}{7}\right)} = 2\sin\left(\frac{k\pi}{7}\right) \]

### Step 3: Calculate \( \log_{7}|1 - z_k| \)

Now we can express the logarithm:

\[ \log_{7}|1 - z_k| = \log_{7}(2\sin\left(\frac{k\pi}{7}\right)) \]

### Step 4: Sum the logarithms

We need to sum these logarithms:

\[ \sum_{k=1}^{6} \log_{7}|1 - z_k| = \sum_{k=1}^{6} \log_{7}(2\sin\left(\frac{k\pi}{7}\right)) \]

Using the property of logarithms, we can separate the terms:

\[ = \sum_{k=1}^{6} \left(\log_{7}(2) + \log_{7}(\sin\left(\frac{k\pi}{7}\right))\right) \]

This can be simplified as:

\[ = 6\log_{7}(2) + \sum_{k=1}^{6} \log_{7}(\sin\left(\frac{k\pi}{7}\right)) \]

### Step 5: Evaluate \( \sum_{k=1}^{6} \log_{7}(\sin\left(\frac{k\pi}{7}\right)) \)

Using the identity for the product of sine functions:

\[ \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}} \]

For \( n = 7 \):

\[ \prod_{k=1}^{6} \sin\left(\frac{k\pi}{7}\right) = \frac{7}{2^{6}} \]

Thus,

\[ \sum_{k=1}^{6} \log_{7}(\sin\left(\frac{k\pi}{7}\right)) = \log_{7}\left(\prod_{k=1}^{6} \sin\left(\frac{k\pi}{7}\right)\right) = \log_{7}\left(\frac{7}{2^{6}}\right) \]

This can be split as:

\[ = \log_{7}(7) - 6\log_{7}(2) = 1 - 6\log_{7}(2) \]

### Step 6: Combine results

Now, substituting back into our expression:

\[ 6\log_{7}(2) + (1 - 6\log_{7}(2)) = 1 \]

### Final Answer

Thus, the final result is:

\[ \boxed{1} \]