# How is $ai^{bi}$ a real number

complex numbersexponentiationnumber theoryreal numbers

$$ai^{bi}$$ 1 immediately appears as an imaginary number: How could an imaginary to the power of yet another imaginary not result in an imaginary?

But, for example, $$3i^{2i}≈0.12964$$ (truncated).

How is it possible that an imaginary to the power of an imaginary is a real number?

Notes:
1: $$a$$ and $$b$$ have the domain $$\{0∉ℝ\}$$ 2
2: Did I format the domain correctly? I've never used set theory before.

#### Best Answer

The number $$i^{i}$$ is real, which immediately implies the result you're interested in. "Proof" : the complex conjugate of a complex number is obtained by replacing every occurrence of $$i$$ with $$-i$$; so the complex conjugate of $$i^{i}$$ is $$(-i)^{-i}=(-1/i)^{i}=i^{i}$$; but the only way that complex conjugation can give back the same number is if the number is actually real.

Addendum: strictly speaking, raising imaginary numbers to imaginary powers is not well-defined. Consider the following. We have $$i=e^{i\pi/2}=e^{5i\pi/2}.$$ Hence $$i^{i}=e^{-\pi/2}=e^{-5\pi/2},$$ but these two real numbers are clearly different. This is a delicate issue, in that resolving this contradiction requires choosing a "branch" of the complex logarithm.