The expression $i^i$, where $i$ is the imaginary unit defined as $i = \sqrt{-1}$, can be evaluated in terms of complex exponentiation.

Step-by-Step Solution

To solve $i^i$, we use the fact that for any complex number $z$, we can write $z = re^{i\theta}$, where $r$ is the magnitude (or modulus) of $z$ and $\theta$ is its argument (angle in the complex plane).

1. Write $i$ in polar form:

   $i$ can be expressed as: $i = e^{i\frac{\pi}{2}}$ because $i$ is 1 unit away from the origin (magnitude = 1), and its angle with the positive real axis is $\frac{\pi}{2}$.

2. Raise $i$ to the power of $i$:

   Now, substitute $i = e^{i\frac{\pi}{2}}$ into $i^i$: $i^i = \left(e^{i\frac{\pi}{2}}\right)^i$

3. Simplify the exponent:

   Use the property $(e^a)^b = e^{ab}$: $i^i = e^{i \cdot i \cdot \frac{\pi}{2}} = e^{-\frac{\pi}{2}}$

Final Result

The result of $i^i$ is: $i^i = e^{-\frac{\pi}{2}} \approx 0.20788$

Interpretation

Surprisingly, $i^i$ is a real number, approximately equal to $0.20788$.

Would you like more details or have any questions?

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Here are some questions to explore this topic further:

1. How does the magnitude and angle of a complex number affect its powers?
2. What other complex values of $i^i$ exist due to the periodic nature of $e^{i\theta}$?
3. Why does raising an imaginary number to an imaginary power yield a real result in this case?
4. Can other complex numbers, like $(-i)^{i}$, be evaluated similarly?
5. How would $i^{\text{real number}}$ behave in general?

Tip: Understanding the polar form of complex numbers helps simplify powers and roots significantly in complex analysis.