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!

Are you familiar with $e^{ix}=\cos x+i\sin x$? If so, start by putting in $x=\pi/2$. – Gerry Myerson Feb 08 '13 at 02:00

That kind of looks like DeMoivre's Theorem, but what exactly happens? – joejacobz Feb 08 '13 at 02:02

joe, why not try it, and see? – Gerry Myerson Feb 08 '13 at 02:04

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. – Julien Feb 08 '13 at 02:17
4 Answers
$$i^i = e^{i\log i} = e^{i(\log i+i\arg i)} = e^{i(i\arg i)} = e^{\arg i} = e^{\frac{\pi}{2}+2 \pi k} \qquad k \in \mathbb{Z}$$
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You seem to say that $\arg i=\pi/2+2\pi ik$. I think it is rather $\pi/2+2\pi k$. – Julien Feb 08 '13 at 02:23

I have always seen everywhere $\arg (re^{i\theta})=\theta +2k\pi$ (when $r>0$ and $\theta\in\mathbb{R}$). – Julien Feb 08 '13 at 02:27
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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4I 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 08 '13 at 02:03


One must be careful about complex powers; this is a branchofthelog consideration. – ncmathsadist Feb 08 '13 at 02:06

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 $n=0$. – guest196883 Feb 08 '13 at 02:08
For example, $$i^i=(\cos(\pi/2)+i\sin(\pi/2))^i=e^{i(i\pi/2)}=e^{\pi/2}$$
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It is not welldefined. 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 08 '13 at 02:41

@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 08 '13 at 02:45

A periodic function is not 11. A function must by 11 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 08 '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 08 '13 at 02:24

@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 08 '13 at 02:47