The value of `i^(i)`, is

Options

A. `-pi/2`
B. `e^(-pi/2)`
C. `e^(pi/2)`
D. none of these

Explanation

To find the value of \( i^i \), we can follow these steps:

### Step 1: Define the expression
Let \( a = i^i \).

### Step 2: Take the logarithm of both sides
Taking the natural logarithm on both sides gives us:
\[ \log a = \log(i^i) \]

### Step 3: Use the power property of logarithms
Using the property of logarithms that states \( \log(m^n) = n \log(m) \), we can rewrite the equation:
\[ \log a = i \log(i) \]

### Step 4: Find \( \log(i) \)
To find \( \log(i) \), we express \( i \) in polar form. The modulus of \( i \) is 1 and the argument (angle) is \( \frac{\pi}{2} \). Therefore:
\[ \log(i) = \log(1) + i \frac{\pi}{2} \]
Since \( \log(1) = 0 \), we have:
\[ \log(i) = i \frac{\pi}{2} \]

### Step 5: Substitute back into the equation
Now substituting \( \log(i) \) back into our equation:
\[ \log a = i \left(i \frac{\pi}{2}\right) \]
This simplifies to:
\[ \log a = -\frac{\pi}{2} \]

### Step 6: Exponentiate to solve for \( a \)
To find \( a \), we exponentiate both sides:
\[ a = e^{-\frac{\pi}{2}} \]

### Final answer
Thus, the value of \( i^i \) is:
\[ i^i = e^{-\frac{\pi}{2}} \]

---