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?

$$i = \sqrt{-1}$$

 

[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]

 

[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]

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.

 

[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.

 

[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.

There must be something wrong with the server, or my browser is working funkily. I somehow cannot view the LaTeX image here...

 

[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.

 

[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)

[tex] \sqrt{i}=i^{\frac{1}{i}}=e^{\frac{\pi}{2}+2n\pi} , n\in\mathbb{Z} [/tex]

Daniel.