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 \(i\), the imaginary unit. For those who don't, \(i=\sqrt{-1}\). But what is \(i^i\)? 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
7 years, 7 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

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.


Daniel Liu - 7 years, 7 months ago

Log in to reply

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 - 7 years, 7 months ago

Log in to reply

Good job!Keep these kind of posts coming!

Bogdan Simeonov - 7 years, 7 months ago

Log in to reply

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

Calvin Lin Staff - 6 years, 9 months ago

Log in to reply

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 - 5 years, 3 months ago

Log in to reply

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

Oskar Limka - 3 years, 1 month ago

Log in to reply

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 - 7 years, 4 months ago

Log in to reply

Yeah it is principal root

Ashish Menon - 5 years, 2 months ago

Log in to reply

What is the square root of -4

Swayam Srivastava - 4 years, 4 months ago

Log in to reply


Tim Lam - 3 years, 6 months ago

Log in to reply

Also -2i

Sonal M - 2 years ago

Log in to reply

@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 years ago

Log in to reply

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

Sonal M - 2 years ago

Log in to reply

What is the value of ixi

Ramkrishna Bojja - 3 years, 8 months ago

Log in to reply


Anatoly Wein - 3 years, 5 months ago

Log in to reply

wait ixi = i^2(x) = -1x = -x

Shayaan Uddin - 7 months ago

Log in to reply

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

Anatoly Wein - 3 years, 5 months ago

Log in to reply

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

Ashish Menon - 3 years, 5 months ago

Log in to reply

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 - 1 year, 11 months ago

Log in to reply

log(i) has a value of approximately 0.68i, which would give: x = i^i log(x) = ilog(i) = ii*0.68 = -0.68 x = 10^(-0.68) = approximately 0.2079

Note: log is base 10, while complex numbers usually are written with base e; which would be ln

Martin Ditlefsen - 1 year, 9 months ago

Log in to reply

What's missing in this presentation is a reminder what seems like a contradiction in the rules of exponentiation. Since all the species in the exponential form of i are being raised by i, that, by real-number doctrine instructs that all species in the exponent are multiplied by i. That would make the new expression e^(-pi/2)*i. That, in turn, if re-expressed in rectangular format is -i.

The only rationale I can find for avoiding this outcome is to invoke DeMoivre's formula, where only the real number value (pi/2) is not addressed by the exponent, i. So, pi/2 remains pi/2 and i becomes i-squared or simply -1. This converts a complex number to real number, ..207 and all its cousins.

Robert Carnes - 1 year, 2 months ago

Log in to reply

What is \cis \cis ?

. . - 5 months ago

Log in to reply

so, what is the modulus and argument?

zigzag shammo - 3 years, 2 months ago

Log in to reply

The argument of a complex number is the degrees (in theta) that it has when written in polar form. The modulus of a complex number is Sqrt(Re(z) ^2 + Im(z) ^2), or for any complex number a+bi, the modulus equals the square root of (a^2 + b^2). The modulus of a complex number z can be written as |z|. It's sort of like the magnitude of z, or the distance from z to the origin, when graphed on the complex plane.

Shayaan Uddin - 7 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...