Question

The value of \(i^i\), (\(i^2 = -1\)) is

• A. 0
• B. 1
• C. -1
• D. \(e^{-\pi/2}\)

Hint:

It is a surprising mathematical fact that an imaginary number raised to an imaginary power (\( i^i \)) results in a purely real number:
\[ i^i = e^{-\pi/2} \approx 0.2079 \]

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question deals with complex analysis and complex numbers.
Specifically, it requires finding the value of an imaginary number raised to an imaginary power (\( i^i \)) using Euler's formula.

Step 2: Detailed Explanation:
Let us represent the imaginary unit \( i \) in exponential (polar) form.
Euler's formula states that for any real number \( \theta \):
\[ e^{i\theta} = \cos \theta + i \sin \theta \] We want to represent \( i \) in this form.
On the complex plane, the number \( i \) lies on the positive imaginary axis, which corresponds to an angle of \( 90^\circ \) or \( \pi/2 \) radians.
Substituting \( \theta = \pi/2 \) into Euler's formula:
\[ e^{i\pi/2} = \cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right) \] Since \( \cos(\pi/2) = 0 \) and \( \sin(\pi/2) = 1 \):
\[ e^{i\pi/2} = 0 + i(1) = i \] Now, let us raise both sides of this exponential representation to the power of \( i \):
\[ i^i = \left( e^{i\pi/2} \right)^i \] Using the laws of exponents, we multiply the powers:
\[ i^i = e^{i \cdot \left(i\frac{\pi}{2}\right)} \] \[ i^i = e^{i^2 \frac{\pi}{2}} \] Since \( i^2 = -1 \), substitute this value:
\[ i^i = e^{-\pi/2} \] Thus, the value of \( i^i \) is a purely real number equal to \( e^{-\pi/2} \).

Step 3: Final Answer:
The value of \( i^i \) is \( e^{-\pi/2} \).
Therefore, the correct choice is Option (D).