Finding the $i$th power of $i$

Question

I have been trying to find the $i$th power of $i$ as a mental exercise, I have tried two approaches

For the first, using properties of exponents

$$i^i=i^{\sqrt{-1}}=i^{-1^{\frac{1}{2}}} \implies i^i=i^{-\frac{1}{2}}=\frac{1}{i^2}=-1$$

However this seems a bit, for a lack of a better word "bodgy".

The next method I used was rewriting into exponential form

$$z=i^i + 0 \implies \arg(z)=\tan^-{1}\left(\frac{1^i}{0}\right)=\frac{\pi}{2}$$
$$ \therefore i^i=e^{\frac{\pi}{2}i}$$

I am quite sure that none of these answers are correct, what are the flaws in my method and how should I go about actually computing this ?

Answer 1

The typical approach to the problem is this: $$ i = e^{\pi i /2} \implies i^i = [e^{\pi i/2}]^i = e^{\pi i^2 /2} = e^{-\pi/2}. $$ However, as I note in my post here, this actually leads to several (infinitely many) possible values for $i^i$. Interestingly, each of these possibilities is a positive real number.

Answer 2

Write

$$i^i = \left(e^{\log i}\right)^i = e^{i \log i}$$

and use the fact that

$$ \log w = \mathrm{ln}|w| + i \,\mathrm{arg}\,w$$

Note that $\mathrm{arg}\,w$ is a multivalued function.