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This question already has an answer here:

i is an imaginary number.

What is $i^i$?

I tried to use euler rule but the answer is strange.

For example $i = e^{\frac{1}{2}i\pi}$.

Using $(a^b)^c = a^{b*c}$ we got

$i^i=e^{(\frac{1}{2}i\pi)*i} = e ^ {-\frac{1}{2}\pi}$

Doesn't seem right

The reason why it doesn't seem right is because of the following

  1. Power by imaginary numbers do not change the size of the result. e^5, is bigger than e^1. However, |e^1000000i| is still just 1.

So that's why I feel it isn't right. So what's the right answer?

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marked as duplicate by Zev Chonoles, Eric Stucky, Dietrich Burde, Asaf Karagila, Did Jun 21 '15 at 8:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Missing a minus sign, but that's the right idea. Hard to believe it's a real number but yeah, it is :P $\endgroup$ – Eric Stucky Jun 21 '15 at 8:35
  • $\begingroup$ @ZevChonoles Yep, I agree with you $\endgroup$ – Ronald Jun 21 '15 at 8:37
  • $\begingroup$ @EricStucky I edited the question to make it better readable and added the minus sign. BTW the answer ($e^{-\frac{1}{2}\pi}$) seems perfectly valid to me $\endgroup$ – Ronald Jun 21 '15 at 8:39
  • $\begingroup$ @EricStucky Actually $i^i$ is not a real number, it is either undefined or a subset of the real line. $\endgroup$ – Did Jun 21 '15 at 8:49
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Under the principal branch of the log. that is $\arg z \in (-\pi, \pi]$ we have that:

$$i^i = e^{i\ln i} = e^{- \pi/ 2} $$

because $\ln z = \ln |z| + i \arg z$. Therefore for $z=i$ we get that $\ln i = \frac{i\pi}{2}$ since $\arg i = \frac{\pi}{2}$. (that is the principal angle) This matches your result. It might be strange at first , but yes it is a real number under the principal branch.

Under a different branch you'll get a different result.

Edit: In more general it holds that:

$$i^{i} = e^{-\pi/2 + 2k \pi}, \;\; k \in \mathbb{Z}$$

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