My $ (i^i)^i $ result does not match Wolfram Alpha's result

Question

See i^i on Wolfram Alpha. It provides a multivalued result that looks like this:

$$ i^i = e^{-2 \pi n - \frac{\pi}{2}} \text{ for } n \in \mathbb{Z} $$

I was able to prove this result. From this result, I try finding the value of $ {(i^i)}^i $. I thought it should be easy.

$$ (i^i)^i = e^{-i2 \pi n - \frac{i\pi}{2}} \text{ for } n \in \mathbb{Z} $$

But see i^i^i on Wolfram Alpha. It provides a multivalued result that looks like this:

$$ c = e^{-2 \pi n - \frac{i \pi}{2}} \text{ for } n \in \mathbb{Z} $$

There is an $ i $ missing in the exponent of $ e $. Is my result incorrect or is Wolfram Alpha's result incorrect or are they both equivalent results and if so how?

Answer 1

By definition, $$(i^i)^i = \exp(i \log( i^i))$$ where $\log$ is any branch of the logarithm. Now $$i^i = \exp( i \log(i)) = \exp(i ( \pi i/2 + 2 n \pi i)) = \exp(-2n\pi - \pi/2)$$ as you said, where $n$ is an arbitrary integer, so $$\log(i^i) = -2n\pi - \pi/2 + 2 m \pi i$$

where both $m$ and $n$ are arbitrary integers, and thus $$(i^i)^i = \exp(-2n\pi i - \pi i/2 - 2 m \pi) = \exp(-\pi i/2 - 2 m \pi) = -i \exp(-2 m \pi)$$

which agrees with Wolfram Alpha.

Answer 2

If $z \neq 0$ and $w$ are complex numbers, then $z^w$ is defined by the formula

$$ z^w := \exp(w \log z)$$

however, the complex logarithm is a multivalued function:

$$ \log z = \ln |z| + i \arg z + 2 \pi i k, \quad \text{for } k \in \mathbb{Z}$$

Note that this means that for complex numbers $x, w$ that $(e^x)^w$ is actually multivalued, with $e^{xw}$ being only one of many answers:

$$ (e^x)^w = \exp(w \log e^x) = \exp(w(x + 2 \pi i k)) = e^{wx} e^{2 w \pi i k}$$

This last $e^{2 w \pi i k}$ for $k \in \mathbb{Z}$ is all that you've missed.