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What is $ i^i$, where $i$ is the imaginary unit. Apparently wolfram alpha and google give: $$i^i\approx0.207879576=e^{-\pi/2}$$ But how?

Maybe let me try: $$x=i^i\implies x=\exp(i\ln i)$$ $$x=\exp\left(i\left(i-\frac{i^2}2+\frac{i^3}3-\frac{i^4}4+\cdots\right)\right)$$ $$x=\exp\left(i\left(i-\frac{1}2-\frac{i}3+\frac{1}4+\cdots\right)\right)$$ $$x=\exp\left(-1-\frac{i}2+\frac{1}3+\frac{i}4+\cdots\right)=^{\text {?}}\exp(-\pi/2)$$

RE60K
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  • Write $i$ in polar form, $i=e^{i\pi/2}$, so $i^i=\big(e^{i\pi/2}\big)^i$. –  Aug 31 '14 at 06:51
  • $\log(i)=\frac{i \pi }{2}$ gives the result $i^i=e^{-\pi /2}$ – Claude Leibovici Aug 31 '14 at 06:56
  • The notation $i^i$ is illegal, as explained in several recent posts on the site. So the answer to the question in the title is "Nonexistent". – Did Aug 31 '14 at 07:21
  • @Did can you specify some? or even one? Soory I didn't paid so attention to all posts. Also it would be better if you can, not necessarily explain, WHY? – RE60K Aug 31 '14 at 07:23
  • An easy argument: $i=e^{5i\pi/2}$ "hence" $i^i=e^{-5\pi/2}$. And $i^i=e^{-9\pi/2}$. And... // The flaw in the approach in your post is that it uses the series expansion of the logarithm at a point where it diverges absolutely. – Did Aug 31 '14 at 07:30
  • @Did: When did "depends on choice of branch cut" become "is illegal"? Have we criminalized the complex logarithm as well? –  Aug 31 '14 at 07:34
  • @Rahul Sorry, if you decide to come back to a more civilized mode of communication, I might explain to you the problem with this notation (if ever my last comment, say, is not enough). But until then... – Did Aug 31 '14 at 07:36
  • @Did: No, I understand what you're saying in your last comment, I think, but it sure seems like exactly the same argument would eliminate notation like $x^{1/2}$ even for positive reals as well. (Eliminate, cause to be illegal (your words), criminalize (= "make illegal", my paraphrase), same thing.) –  Aug 31 '14 at 07:46
  • @Rahul This is a strawman argument. Mentioning the (serious) problem with $z^{1/2}$ for $z$ complex is not the same as saying there would be a problem with $x^{1/2}$ for $x$ real nonnegative (there is none). – Did Aug 31 '14 at 08:08
  • I remember when my teacher said this was impossible, if it weren't for $cis \theta$ , I wouldn't ever have proved him wrong. – Nick Aug 31 '14 at 09:08
  • @Nick Yes... That such a banal piece of maths is facing a furious opposition, repeatedly, and from several math.se users, is something worthy of an explanation. This might be related to some specific educational systems deliberately neglecting the point (and even, misleading students about it). Do you have any informed view about this? – Did Aug 31 '14 at 11:19
  • @Did: This is the Mathematician vs Math-teacher phenomenon. There's just too far a gap between the two that the latter doesn't even need to know how to count if he knows the answers to the questions in the textbook. So, that's what I think the primary cause is. – Nick Aug 31 '14 at 11:50

1 Answers1

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Writing $i = \exp\left(\frac{\pi}{2} i\right)$, we have

$i^i = \exp\left(\frac{\pi}{2} i\cdot i\right) = e^{-\pi / 2}$

as desired.

Travis Willse
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  • Does ?$i=\exp(i\pi/2)=\cos(\pi/2)+i\sin(\pi/2)=i$, then $i^i$ would be $e^{i\ln i}=e^{i\ln(e^{i\pi/2})}=e^{i.i\pi/2}=e^{-\pi/2}$? – RE60K Aug 31 '14 at 07:04
  • @Aditya yes. ${} {}$ – Dustan Levenstein Aug 31 '14 at 07:21
  • @Aditya Yes, that's right, but note that, as your computation makes explicit this computation depends implicitly on a choice of branch cut for $\ln$; more explicitly, to write $i^i = e^{-\pi/2}$, we must use a branch cut for which $\ln i = \frac{\pi}{2} i$, and the "usual" does just find. If this sort of situation bothers you, just write your expression in the form $\exp w$ from the outside. – Travis Willse Aug 31 '14 at 08:23