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

[Reshma]

A small cute question here .

i[/tex] is imaginary. So what is the nature of i^i. Is it imaginary too? <br /> <br /> 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:
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:
i^{i}=e^{i*(i*(\frac{\pi}{2}+2k\pi))}=e^{-\frac{\pi}{2}-2k\pi}[/itex]

<br /> <br /> All of which are, by the way, real, not &quot;pure imaginary&quot;!

 

[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 i^i[/b], Reshma said &quot;How is it possible to show that? Isn&#039;t i^i purely imaginary?&quot; so I thought it best to make it very clear that all &quot;variations&quot; 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 i^i[/b], Reshma said &quot;How is it possible to show that? Isn&#039;t i^i purely imaginary?&quot; so I thought it best to make it very clear that all &quot;variations&quot; 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 i^i[/b], Reshma said &quot;How is it possible to show that? Isn&#039;t i^i purely imaginary?&quot; so I thought it best to make it very clear that all &quot;variations&quot; of iⁱ were real, not imaginary.

<br /> There must be something wrong with the server, or my browser is working funkily. I somehow cannot view the LaTeX image here... <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":confused:" title="Confused :confused:" data-smilie="5"data-shortname=":confused:" />

 

[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 i^i[/b], Reshma said &quot;How is it possible to show that? Isn&#039;t i^i purely imaginary?&quot; so I thought it best to make it very clear that all &quot;variations&quot; 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)

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

Daniel.