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My query is about the $i^{i}$ , where $i$ is defined to be the imaginary unit, and $i \in C$.

I know the proof of this value, we just have to substitute $i$ as $\large{e^{i\left(\frac{\pi}{2}+2n\pi\right)}}$. Where $n$ is any integer because all these values corresponds to "$i$" right?

Which gives us, $i^{i}=\large{e^{-\frac{\pi}{2}+2n\pi}}$ Where $n=0$, is the principal value. $i.e.,$ $e^{-\frac{\pi}{2}}$.

I mean this result always blow my mind, to be honestly speaking. Not only that $i^i$ despite being a complex number raised to a complex number. It also attains a real value! And that too not just one, but many!

My question would be that first, is this result strictly valid in accordance to mathematics? Are there any loopholes in defining it as such? Here's what I think it could be of importance http://www.cut-the-knot.org/do_you_know/complex.shtml as some quotes in this page indicates.

Secondly, is there any intuition behind why could it be so, even if it could, the idea that it can take infinitely many values on $\mathcal R$ just by changing "$n$" is still hard kind of.

Someone
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    consider just the simple case, $z^{1/2}$, we see, when $z$ is a complex number, that this "function" has two values for every $z$. In general $z^w$ has a single value only when $w$ is an integer. It has exactly $q$ values when $w=p/q$ is the reduced reprenstation of a rational number. In any other case, it has infinitely many values. – Thomas Andrews May 20 '15 at 18:09
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    So $1^i$ even takes on infinitely many values. – Thomas Andrews May 20 '15 at 18:11
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    For your first question: indeed, $i = e^{i (\pi/2 + 2n\pi)}$ is a direct consequence of Euler's formula $e^{ix} = \cos(x) + i \sin(x)$. – A.P. May 20 '15 at 18:13
  • This question is very nicely discussed on this site, link'this' – Cloverr May 21 '15 at 18:48

1 Answers1

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I know that I added a "contradictory" comment as you will see, but a "rigorous" intuition will be,
$i^i=(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})^i$
Now, by De Moivre's Theorem,
$i^i=\cos i\frac{\pi}{2}+i\sin i\frac{\pi}{2}$
By thee Euler's Formula,
$i^i=e^{i^2\frac{\pi}{2}}$
$i^i=e^{-\frac{\pi}{2}}$
$i^i=\frac{1}{e^\frac{\pi}{2}}$
Which, according to my calculator is, $0.207879576...$.
Which is confirmed by google.
P.S.:I didn't prove it for $2n\pi$, just a case of it.
As far as you want the intuition, think the reverse, isn't it normal for the complex root of a real number to be complex? For example, $2^{\frac1i}$ is complex. So, the question you are asking is just the reverse of it.
I hope it makes sense.

Aditya Agarwal
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