What is the Nature of i to the Power of i? Understanding Imaginary Numbers

[Reshma]

A small cute question here .

[itex]i[/tex] is imaginary. So what is the nature of $i^i$. Is it imaginary too? <br /> <br /> $$i = \sqrt{-1}$$[/itex]

 

[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.

 

[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.

 

[Reshma]

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.

How is it possible to show that? Isn't $i^i$ purely imaginary?

 

[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:
[tex]i^{i}=e^{i*(i*(\frac{\pi}{2}+2k\pi))}=e^{-\frac{\pi}{2}-2k\pi}[/itex][/tex]

 

[HallsofIvy]

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:
[tex]i^{i}=e^{i*(i*(\frac{\pi}{2}+2k\pi))}=e^{-\frac{\pi}{2}-2k\pi}[/itex][/tex]

$$<br /> All of which are, by the way, real, not "pure imaginary"!$$

 

[arildno]

Can't see I said they were imaginary, but I should have emphasized them being real nonetheless.
Thanks, HallsofIvy

 

[HallsofIvy]

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 [itex]i^i[/b], Reshma said "How is it possible to show that? Isn't $i^i$ purely imaginary?" so I thought it best to make it very clear that all "variations" of iⁱ were real, not imaginary.[/itex]

 

[HallsofIvy]

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 [itex]i^i[/b], Reshma said "How is it possible to show that? Isn't $i^i$ purely imaginary?" so I thought it best to make it very clear that all "variations" of iⁱ were real, not imaginary.[/itex]

 

[arildno]

I fully agree. I ought to have emphasized it, your clarification was necessary.

 

[VietDao29]

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 [itex]i^i[/b], Reshma said "How is it possible to show that? Isn't $i^i$ purely imaginary?" so I thought it best to make it very clear that all "variations" of iⁱ were real, not imaginary.[/itex]

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[HallsofIvy]

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 [itex]i^i[/b], Reshma said "How is it possible to show that? Isn't $i^i$ purely imaginary?" so I thought it best to make it very clear that all "variations" of iⁱ were real, not imaginary.[/itex]

 

[Reshma]

Thank you very much, Arildno and HallsofIvy! So, $i^i$ is real !

 

[dextercioby]

The same goes for the more spectacular (when it comes to notation)

$$\sqrt<i>{i}=i^{\frac{1}{i}}=e^{\frac{\pi}{2}+2n\pi} , n\in\mathbb{Z} </i>$$

Daniel.