lf `z = ilog(2-sqrt(3))`, then `cos z =`

Options

A. `-1`
B. `-1//2`
C. 1
D. 2

Explanation

To solve the problem where \( z = i \log(2 - \sqrt{3}) \) and we need to find \( \cos z \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the expression for \( z \)**:
\[ z = i \log(2 - \sqrt{3}) \]

2. **Use Euler's formula**:
Recall that Euler's formula states:
\[ e^{iz} = \cos(z) + i\sin(z) \]
From this, we can express \( \cos(z) \) as:
\[ \cos(z) = \frac{e^{iz} + e^{-iz}}{2} \]

3. **Calculate \( e^{iz} \)**:
Substitute \( z \) into the expression:
\[ e^{iz} = e^{i(i \log(2 - \sqrt{3}))} = e^{-\log(2 - \sqrt{3})} \]
Using the property of logarithms, we have:
\[ e^{-\log(a)} = \frac{1}{a} \]
Therefore:
\[ e^{iz} = \frac{1}{2 - \sqrt{3}} \]

4. **Calculate \( e^{-iz} \)**:
Similarly, we find \( e^{-iz} \):
\[ e^{-iz} = e^{-\left(-i \log(2 - \sqrt{3})\right)} = e^{i \log(2 - \sqrt{3})} = 2 - \sqrt{3} \]

5. **Combine the results**:
Now we can substitute \( e^{iz} \) and \( e^{-iz} \) back into the formula for \( \cos(z) \):
\[ \cos(z) = \frac{e^{iz} + e^{-iz}}{2} = \frac{\frac{1}{2 - \sqrt{3}} + (2 - \sqrt{3})}{2} \]

6. **Simplify the expression**:
To simplify \( \frac{1}{2 - \sqrt{3}} + (2 - \sqrt{3}) \), we first rationalize \( \frac{1}{2 - \sqrt{3}} \):
\[ \frac{1}{2 - \sqrt{3}} \cdot \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3} \]
Thus:
\[ \cos(z) = \frac{(2 + \sqrt{3}) + (2 - \sqrt{3})}{2} = \frac{4}{2} = 2 \]

7. **Final result**:
Therefore, the value of \( \cos(z) \) is:
\[ \cos(z) = 2 \]