The expression \(i^i\), where \(i\) is the imaginary unit (defined as \(i = \sqrt{-1}\)), can be evaluated using Euler's formula:

\[
e^{ix} = \cos(x) + i\sin(x)
\]

Let's rewrite \(i\) in exponential form:
\[
i = e^{i\frac{\pi}{2}}
\]

Now, raise \(i\) to the power of \(i\):
\[
i^i = \left(e^{i\frac{\pi}{2}}\right)^i
\]

This simplifies to:
\[
i^i = e^{i \cdot i\frac{\pi}{2}} = e^{-\frac{\pi}{2}}
\]

Thus, \(i^i\) is a real number:
\[
i^i = e^{-\frac{\pi}{2}} \approx 0.20788
\]

This is the principal value of \(i^i\), but there are infinitely many values due to the periodic nature of the exponential function involving complex numbers.

Would you like more details or have any questions?

Here are 5 related questions to explore further:
1. What are the possible values of \(i^i\) due to the periodicity of the exponential function?
2. How does Euler's formula help in simplifying complex number expressions?
3. Can \(i^i\) be expressed in terms of trigonometric functions?
4. What is the geometric interpretation of raising \(i\) to a power?
5. How do complex logarithms relate to the different values of \(i^i\)?

**Tip:** Remember that complex exponentiation can yield multiple values due to the multi-valued nature of the logarithm function in the complex plane.

Explore the evaluation of i^i, where i is the imaginary unit. Learn how Euler's formula simplifies complex exponentiation, and discover the principal value of i^i as a real number. Dive into the periodic nature of exponential functions in complex numbers.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation