The real part of the principle value of ` 2^( - i)` is

Options

A. ` sin ( log 2 ) `
B. ` cos ((1 )/( log 2 )) `
C. ` cos [ log ((1 )/( 2 )) ] `
D. ` cos ( log 2 ) `

Explanation

To find the real part of the principal value of \( 2^{-i} \), we can follow these steps:

### Step 1: Rewrite the expression
Let \( z = 2^{-i} \).

### Step 2: Use logarithmic properties
We can express this using the exponential form:
\[ z = e^{-i \log(2)} \]
Here, we use the property that \( a^b = e^{b \log(a)} \).

### Step 3: Apply Euler's formula
Using Euler's formula \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \), we can rewrite \( z \):
\[ z = e^{-i \log(2)} = \cos(-\log(2)) + i\sin(-\log(2)) \]

### Step 4: Simplify the expression
Since \( \cos(-x) = \cos(x) \) and \( \sin(-x) = -\sin(x) \), we have:
\[ z = \cos(\log(2)) - i\sin(\log(2)) \]

### Step 5: Identify the real part
The real part of \( z \) is:
\[ \text{Re}(z) = \cos(\log(2)) \]

### Final Answer
Thus, the real part of the principal value of \( 2^{-i} \) is:
\[ \cos(\log(2)) \]

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