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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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
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That kind of looks like DeMoivre's Theorem, but what exactly happens? – joejacobz Feb 08 '13 at 02:02
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joe, why not try it, and see? – Gerry Myerson Feb 08 '13 at 02:04
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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
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$$i^i = e^{i\log i} = e^{i(\log |i|+i\arg i)} = e^{i(i\arg i)} = e^{-\arg i} = e^{-\frac{\pi}{2}+2 \pi k} \qquad k \in \mathbb{Z}$$
Argon
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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
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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
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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.
guest196883
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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
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One must be careful about complex powers; this is a branch-of-the-log consideration. – ncmathsadist Feb 08 '13 at 02:06
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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
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For example, $$i^i=(\cos(\pi/2)+i\sin(\pi/2))^i=e^{i(i\pi/2)}=e^{-\pi/2}$$
Clayton
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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 08 '13 at 02:41
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@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
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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 08 '13 at 13:59
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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 $$i^i=(e^{\frac{\pi i}{2}})^i=e^{\frac{-\pi }{2}}.$$
i.a.m
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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
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@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