What is i to the power of i?

Disclaimer: knowledge of complex numbers and trigonometry is needed to fully appreciate this note.

Some of you may know the definition of ii, the imaginary unit. For those who don't, i=1i=\sqrt{-1}. But what is iii^i? Is it an imaginary number? Perhaps even one step deeper than imaginary, a quaternary number?

In fact, iii^i is a real number! It approximately equals 0.207880.20788.

But how can that be? How can an imaginary number to the power of an imaginary number be real? How did I arrive at that inexplicable answer?

To find the value of iii^i, we'll first need to know Euler's Formula.

Euler's Formula states that eix=cosx+isinxe^{ix}=\cos{x}+i\sin{x}. The right hand side can be shortened to cis x\text{cis }{x} (just notation, don't get confused), and I'll be using this shorthand for brevity.

Note that when x=π2x=\dfrac{\pi}{2}, then cis x=cosπ2+isinπ2    cis x=i\text{cis }{x}=\cos{\dfrac{\pi}{2}}+i\sin{\dfrac{\pi}{2}}\implies\text{cis }{x}=i. Therefore, to find iii^i, we can simply take cis π2\text{cis }{\dfrac{\pi}{2}} to the iith power. However, since cis π2=eiπ2\text{cis }{\dfrac{\pi}{2}}=e^{i\frac{\pi}{2}}, then

ii=(cis π2)i=ei2π2=eπ20.20788\begin{aligned}i^i&=(\text{cis }{\dfrac{\pi}{2}})^i\\ &=e^{i^2\frac{\pi}{2}}\\ &=e^{-\frac{\pi}{2}}\\ &\approx 0.20788\end{aligned} and we are done.

However, the astute reader would notice something wrong with my argument. When I said x=π2x=\dfrac{\pi}{2} gives cis x=i\text{cis }{x}=i, I was only considering one scenario of the infinite possible xx that satisfy cis x=i\text{cis }{x}=i. What about x=5π2x=\dfrac{5\pi}{2}? How about x=3π2x=\dfrac{-3\pi}{2}? Any xx satisfying x=π2+2πnx=\dfrac{\pi}{2}+2\pi n for some integer nn would work, and every single nn gives a different value for iii^i. Which one is the correct answer?

The short answer is: all of them are equally correct. But how can that be? Shouldn't iii^i, a constant to the power of a constant, yield only one value? One reason that iii^i might not give what is expected is the fact that it is to the power of ii. What does it mean to take an iith power? The fundamental meaning of taking "to the power" breaks down when you consider this.

You can make this weird fact more acceptable to yourself by asking yourself: What are all the possible values of 4124^{\frac{1}{2}}? There are, in fact, two values: 2-2 and 22.

Therefore along those lines of reasoning, ii=0.20788i^i=0.20788, given by n=0n=0. Also, ii=111.31778i^i=111.31778, where n=1n=-1. ii=0.00039i^i=0.00039, where n=1n=1.

But much like the principle value of 412=24^{\frac{1}{2}}=2, we also have the principle value of iii^i, which is when n=0n=0. So, finally, we have shown that iii^i does indeed equal 0.207880.20788. \Box

Got any questions? Feel free to ask below.

Note by Daniel Liu
5 years, 10 months ago

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Thanks for reading through my second #CosinesGroup post, this time about imaginary powers (specifically iii^i). I hope you all found it quite intriguing and informative. If you have any questions or feedback, feel free to post.

Cheers!

Daniel Liu - 5 years, 10 months ago

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I know this probably isn't fit for #CosinesGroup, but I'll post it anyways. Just recall that for complex x,y x,y , we can define xy=exp(ylogx) x^y = \exp(y \log x) . Also, remember that the complex logarithm is defined as logx=logx+iargx \log x = \log|x| + i \arg|x| . So if we fix the range of the argument then we can perfectly well define xy x^y as we wish.

Just sub in the values of x,y=i x,y = i to get:

ii=exp(argi) i^i = \exp(-\arg i) .

And we are done.

Anqi Li - 5 years, 10 months ago

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Good job!Keep these kind of posts coming!

Bogdan Simeonov - 5 years, 10 months ago

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@Daniel Liu Can you add this to a suitable skill in the Complex Numbers Wiki? Thanks

Calvin Lin Staff - 5 years ago

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Thank you very much..amazing note..I was discussing this concept with my friends in my coaching classes..now I understand it completely..thank u sir..@Daniel Liu

Rishabh Tiwari - 3 years, 5 months ago

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Great post. You get an A+ for the math and A- for English: it's "principal" (not "principle") value ;-)

Oskar Limka - 1 year, 3 months ago

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I don't think that √4 is equal to both 2 & -2 Rather its actually just 2 since √x is always pos. For real x.

Otherwise , nice thoughts !!!!

Vishal Sharma - 5 years, 6 months ago

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Yeah it is principal root

Ashish Siva - 3 years, 5 months ago

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What is the square root of -4

Swayam Srivastava - 2 years, 6 months ago

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2i

Tim Lam - 1 year, 8 months ago

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Also -2i

Sonal M - 2 months, 3 weeks ago

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@Sonal M 2i is the principle root. If you really wanted you could pick 2i^6 and so you might as well list out all the potential answers.

Tim Lam - 2 months, 3 weeks ago

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@Tim Lam 2i^6=2 * i^4 * i^2=2 * 1 * -1=-2

Sonal M - 2 months, 3 weeks ago

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What is the value of ixi

Ramkrishna Bojja - 1 year, 11 months ago

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-1

Anatoly Wein - 1 year, 7 months ago

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e^pi ~ 9 if we take the i-th power we get -1 according e^pi^i+1=0. Why?

Anatoly Wein - 1 year, 7 months ago

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eiπ =cos(π)+i×sin(π) =1e^{i\pi} \ = \cos(\pi) + i × \sin(\pi) \ = -1

Ashish Siva - 1 year, 7 months ago

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Log(I) has the value of about 0.68. I’m not a mathematician, and I’m not smart enough too work out how they worked that out.

But unless they are wrong and just making shit up...

Let x = i^i

Log(x) = log(i^i) = i * log(i) = 0.68 * i

Log(log(x)) = log(log(i^i)) = log (0.68 * i) = log(0.68) + log(I) = -0.38 + 0.68

Log(log(i^i)) ~= 0.3

i^i = e^e^0.3 >= 1

Graeme Smith - 2 months ago

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so, what is the modulus and argument?

zigzag shammo - 1 year, 4 months ago

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