# Nature of this number

1. Feb 15, 2006

### Reshma

A small cute question here .

$i[/tex] is imaginary. So what is the nature of $i^i$. Is it imaginary too? $$i = \sqrt{-1}$$ 2. Feb 15, 2006 ### schattenjaeger knowing that i is e^(i*pi/2), raising it to the ith power gives you e^(i^2*pi/2) and i^2 is -1, so e^(-pi/2) *pulls out Ti-89* yup. 3. Feb 15, 2006 ### arildno The question is ill-posed. There isn't just one number that may lay claim to the representation as $i^{i}$; infinitely many numbers, in fact, can claim that right. 4. Feb 15, 2006 ### Reshma How is it possible to show that? Isn't $i^i$ purely imaginary? 5. Feb 15, 2006 ### arildno It has to do with the complex logarithm being non-unique, we have, say for any choice of integer k: $$i=e^{i\frac{\pi}{2}+i2k\pi}$$ Thus, the logarithm of i is the set of numbers $i(\frac{\pi}{2}+2k\pi), k\in{Z}$ Thus, we have: $$i^{i}=e^{i*(i*(\frac{\pi}{2}+2k\pi))}=e^{-\frac{\pi}{2}-2k\pi}$ 6. Feb 15, 2006 ### HallsofIvy Staff Emeritus All of which are, by the way, real, not "pure imaginary"! 7. Feb 15, 2006 ### arildno Can't see I said they were imaginary, but I should have emphasized them being real nonetheless. Thanks, HallsofIvy 8. Feb 15, 2006 ### HallsofIvy Staff Emeritus You didn't. I was emphasizing that they are real since the original post was "$i$ is imaginary. So what is the nature of $i^i$. Is it imaginary too?" and then after you said there were many numbers equal to $i^i[/b], Reshma said "How is it possible to show that? Isn't [itex]i^i$ purely imaginary?" so I thought it best to make it very clear that all "variations" of ii were real, not imaginary. 9. Feb 15, 2006 ### HallsofIvy Staff Emeritus You didn't. I was emphasizing that they are real since the original post was "$i$ is imaginary. So what is the nature of $i^i$. Is it imaginary too?" and then after you said there were many numbers equal to $i^i[/b], Reshma said "How is it possible to show that? Isn't [itex]i^i$ purely imaginary?" so I thought it best to make it very clear that all "variations" of ii were real, not imaginary. 10. Feb 15, 2006 ### arildno I fully agree. I ought to have emphasized it, your clarification was necessary. 11. Feb 15, 2006 ### VietDao29 There must be something wrong with the server, or my browser is working funkily. I somehow cannot view the LaTeX image here... 12. Feb 15, 2006 ### HallsofIvy Staff Emeritus You didn't. I was emphasizing that they are real since the original post was "$i$ is imaginary. So what is the nature of $i^i$. Is it imaginary too?" and then after you said there were many numbers equal to $i^i[/b], Reshma said "How is it possible to show that? Isn't [itex]i^i$ purely imaginary?" so I thought it best to make it very clear that all "variations" of ii were real, not imaginary. 13. Feb 16, 2006 ### Reshma Thank you very much, Arildno and HallsofIvy! So, $i^i$ is real !! 14. Feb 17, 2006 ### dextercioby The same goes for the more spectacular (when it comes to notation) [tex] \sqrt{i}=i^{\frac{1}{i}}=e^{\frac{\pi}{2}+2n\pi} , n\in\mathbb{Z}$$

Daniel.

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