What does it mean to take $i^i$?

Question

Background

My friend derived that $$i^i = e^{-(4m+1) \frac {\pi}{2} }$$ (though he was unaware about such thing earlier!). Initially my reaction was that there might be some flaw in the derivation process but after checking it thoroughly it felt genuine. Now while trying to find more about it I found this article which validates his derivation. Even though now I know it is correct but I don't know know about its mathematical significance.

Elaboration

I am quite comfortable with integral powers but when it comes to non-integral powers it becomes quite tough. For example when we ask what is $2^{1/3}$ it becomes a bit tougher but still it is a bit understandable that we are being asked of a number whose cube is 2. Then comes irrational part though it being much harder is still understandable like when asked about $2^{\pi}$ I can say that it is equal to $$2^3 \times 2^ \frac{1}{10} \times 2^ \frac{4}{100} \times 2^ \frac{1}{1000} \times ...$$. But then comes this $i^i$ which I can't understand even a little bit

Question

So the questions are:

Can it be visualized in some way as is depicted above and if possible can you also explain it with Argand Plane?(Most Important)
What is its mathematical significance?

Thanks in advance.

EDIT

This YouTube video gives a somewhat the answer I expected for this question but to a different question related to the visual interpretation of $e^{\iota \pi} = -1$.

Are you familiar with the general definition $x^y=\exp(y\log x)$? — J. M. ain't a mathematician, Nov 04 '19 at 13:21
If you're thinking of exponents as repeated multiplication, then $i^i$ is indeed nonsensical. You need a different mental model of what exponentiation is. Like the one suggested in the comment above. — Arthur, Nov 04 '19 at 13:22
There are several duplicates discussing exactly what you wanted to know - how to understand $i^i$, right? All the answers explain this, e.g., taking the logarithm resp. exponential etc. Have you seen the answers already? — Dietrich Burde, Nov 04 '19 at 13:27
You mean, where Tolaso answers it? But he answers exactly to $i^i$? — Dietrich Burde, Nov 04 '19 at 13:30

score 1 · Answer 1 · answered Nov 04 '19 at 13:31

All boils down it the following identity

$$e^{i\theta}=\cos \theta + i \sin \theta$$

therefore since $i=e^{i\left(\frac{\pi}2+2m\pi\right)}$ we have

$$i^i= e^{\log i^i}=e^{i\log i}=e^{i\log(e^{i\left(\frac{\pi}2+2m\pi\right)})}=e^{-\frac{\pi}2-2m\pi}$$

Refer also to

How do I interpret Euler's formula?

score 1 · Answer 2 · answered Nov 04 '19 at 13:40

In complex analysis we define $a^b$ to be $\exp(b\ln a)$ where $a,b\in\mathbb{C}$. Therefore, we have $$i^i=\exp(i\ln i)$$ but here you will have to face another issue: how should we define $\ln a$ with $a\in\mathbb{C}$?

Here we must resort to the famous Euler formula that $\exp(i\theta)=\cos\theta+i\sin\theta$. With this magical formula, we can define complex exponent! Then we can define $\ln a$ to be the inverse of $\exp$, which means if $b=\ln a$, we have $a=\exp b$. Then we successfully extend the definition of $\ln$ from positive numbers to the punctured complex plane $\mathbb{C}\backslash\{0\}$ (the origin must be removed since the modulus of $\exp b$ is always positive)!

Notice that unlike the usual real functions we encounter, this definition requires $\ln$ to be multi-valued. Even for real positive numbers, this new definition gives multi-valued results. Consider $a>0$ real, it follows immediately that $\ln a+ 2k\pi i$ satisfies $\exp a=b$ for every integer $k$, where $\ln a$ is the real value in our common sense without complex analysis.

What does it mean to take $i^i$?

2 Answers2