# What is the value of $i^i$? [duplicate]

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 $x=\pi/2$. Feb 8, 2013 at 2:00
• That kind of looks like DeMoivre's Theorem, but what exactly happens? Feb 8, 2013 at 2:02
• joe, why not try it, and see? Feb 8, 2013 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. Feb 8, 2013 at 2:17

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

• There is an extra $i$ in your last term. Feb 8, 2013 at 2:19
• You seem to say that $\arg i=\pi/2+2\pi ik$. I think it is rather $\pi/2+2\pi k$. Feb 8, 2013 at 2:23
• I have always seen everywhere $\arg (re^{i\theta})=\theta +2k\pi$ (when $r>0$ and $\theta\in\mathbb{R}$). Feb 8, 2013 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.

• 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. Feb 8, 2013 at 2:03
• Yes, caveat emptor. Feb 8, 2013 at 2:05
• One must be careful about complex powers; this is a branch-of-the-log consideration. Feb 8, 2013 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 $n=0$. Feb 8, 2013 at 2:08

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

• No......... This does not work. Feb 8, 2013 at 2:05
• @ncmathsadist: Why not? Just curious. Feb 8, 2013 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 $\log$? Feb 8, 2013 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? Feb 8, 2013 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. Feb 8, 2013 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 $$i^i=(e^{\frac{\pi i}{2}})^i=e^{\frac{-\pi }{2}}.$$

• @Argon I know that, but it is still true, the argument of $i$ is $\frac{\pi}{2}$ Feb 8, 2013 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. Feb 8, 2013 at 2:47