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I understand that when you raise any number $x$ to a power, you multiply $x$ by itself the number of times indicated in the power. However, what happens when $i^i$ is performed? How can a number be multiplied an imaginary amount of times? Wolfram Alpha says that it is equal to $e^{{-\pi}/{2}}$, but how would you arrive at that answer? Any response will be appreciated, thanks!

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marked as duplicate by Ross Millikan, Micah, Gerry Myerson, Argon, Asaf Karagila Feb 8 '13 at 2:21

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Are you familiar with $e^{ix}=\cos x+i\sin x$? If so, start by putting in $x=\pi/2$. – Gerry Myerson Feb 8 '13 at 2:00
That kind of looks like DeMoivre's Theorem, but what exactly happens? – joejacobz Feb 8 '13 at 2:02
joe, why not try it, and see? – Gerry Myerson Feb 8 '13 at 2:04
As noted in particular by L.F., Argon and ncmathsadist, the answer depends on the branch of the complex logarithm you choose to work with. – 1015 Feb 8 '13 at 2:17
up vote 5 down vote accepted

$$i^i = e^{i\log i} = e^{i(\log |i|+i\arg i)} = e^{i(i\arg i)} = e^{-\arg i} = e^{-\frac{\pi}{2}+2 \pi k} \qquad k \in \mathbb{Z}$$

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There is an extra $i$ in your last term. – 1015 Feb 8 '13 at 2:19
You seem to say that $\arg i=\pi/2+2\pi ik$. I think it is rather $\pi/2+2\pi k$. – 1015 Feb 8 '13 at 2:23
I have always seen everywhere $\arg (re^{i\theta})=\theta +2k\pi$ (when $r>0$ and $\theta\in\mathbb{R}$). – 1015 Feb 8 '13 at 2:27

Well, in the complex numbers you consider an exponential of a base other than $e$, such as $z^x$, to be: $$z^x := e^{x\log z }$$ So we have: $$i^i = e^{i\log i}$$ But $\log i = i\left(\frac{\pi}{2}+2\pi n\right)$, so we have $$i^i = e^{ii(\frac{\pi}{2}+2\pi n)} = e^{\frac{-\pi}{2} + 2\pi n} ~~~~~~~~~~ n\in \Bbb{Z}$$ Taking $n=0$ gives the value that Wolfram Alpha gave you.

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I think it should be mentioned that $e^{-\frac{\pi}{2}}$ is the principal value of the expression, $i^i$ can take infinitely many real values. – L. F. Feb 8 '13 at 2:03
Yes, caveat emptor. – ncmathsadist Feb 8 '13 at 2:05
One must be careful about complex powers; this is a branch-of-the-log consideration. – ncmathsadist Feb 8 '13 at 2:06
Yes, as L.F. said, log i can produce an infinite number of values, and is only a function upon selecting a branch $i(\frac{\pi}{2} + 2\pi n). $ $e^{-\pi/2}$ is the principal value with $n=0$. – Sam DeHority Feb 8 '13 at 2:08

For example, $$i^i=(\cos(\pi/2)+i\sin(\pi/2))^i=e^{i(i\pi/2)}=e^{-\pi/2}$$

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No......... This does not work. – ncmathsadist Feb 8 '13 at 2:05
@ncmathsadist: Why not? Just curious. – Clayton Feb 8 '13 at 2:37
It is not well-defined. The map $z \mapsto z^i$ needs to be defined as $z \mapsto e^{i\log(z)}$. Since the complex exponential is periodic, what's the meaning of $\log$? – ncmathsadist Feb 8 '13 at 2:41
@ncmathsadist: I see; you're saying it is essentially because $\sin$ is periodic? I mean, we can take $5\pi/2$ and get a different answer? – Clayton Feb 8 '13 at 2:45
A periodic function is not 1-1. A function must by 1-1 to possess an inverse. You achieve this with the trig functions by pruning the domain. The same thing must be done in complex analysis with the log function. – ncmathsadist Feb 8 '13 at 13:59

for any complex number $z\in \mathbb C$ it can be written as $z=x+iy$ or in the polar form $z=re^{i\theta}$ where $r=\sqrt{x^2+y^2}$ and $\theta=\arctan(\frac{y}{x})$, so in particular $i=0+i1$ which implies that $r=1$ and $\theta=\frac{\pi}{2}$ Thus, $$i=e^{\frac{\pi i}{2}}$$ which implies that $$i^i=(e^{\frac{\pi i}{2}})^i=e^{\frac{-\pi }{2}}.$$

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@Argon I know that, but it is still true, the argument of $i$ is $\frac{\pi}{2}$ – i.a.m Feb 8 '13 at 2:24
@Argon I understand what you mean, but may be I should've just written that $\theta=\frac{\pi}{2}$ without trying to explain it more. – i.a.m Feb 8 '13 at 2:47

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