# 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, $i^i$ is a real number! It approximately equals $0.20788$.

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 $i^i$, we'll first need to know Euler's Formula.

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

Note that when $x=\dfrac{\pi}{2}$, then $\text{cis }{x}=\cos{\dfrac{\pi}{2}}+i\sin{\dfrac{\pi}{2}}\implies\text{cis }{x}=i$. Therefore, to find $i^i$, we can simply take $\text{cis }{\dfrac{\pi}{2}}$ to the $i$th power. However, since $\text{cis }{\dfrac{\pi}{2}}=e^{i\frac{\pi}{2}}$, then

\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 $x=\dfrac{\pi}{2}$ gives $\text{cis }{x}=i$, I was only considering one scenario of the infinite possible $x$ that satisfy $\text{cis }{x}=i$. What about $x=\dfrac{5\pi}{2}$? How about $x=\dfrac{-3\pi}{2}$? Any $x$ satisfying $x=\dfrac{\pi}{2}+2\pi n$ for some integer $n$ would work, and every single $n$ gives a different value for $i^i$. Which one is the correct answer?

The short answer is: all of them are equally correct. But how can that be? Shouldn't $i^i$, a constant to the power of a constant, yield only one value? One reason that $i^i$ might not give what is expected is the fact that it is to the power of $i$. What does it mean to take an $i$th 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 $4^{\frac{1}{2}}$? There are, in fact, two values: $-2$ and $2$.

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

But much like the principle value of $4^{\frac{1}{2}}=2$, we also have the principle value of $i^i$, which is when $n=0$. So, finally, we have shown that $i^i$ does indeed equal $0.20788$. $\Box$

Got any questions? Feel free to ask below. Note by Daniel Liu
6 years, 11 months ago

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

Cheers!

- 6 years, 11 months ago

Good job!Keep these kind of posts coming!

- 6 years, 11 months ago

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

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

$i^i = \exp(-\arg i)$.

And we are done.

- 6 years, 11 months ago

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

Staff - 6 years, 1 month ago

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

- 4 years, 7 months ago

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

- 2 years, 5 months ago

What is the square root of -4

- 3 years, 8 months ago

2i

- 2 years, 10 months ago

Also -2i

- 1 year, 4 months ago

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.

- 1 year, 4 months ago

2i^6=2 * i^4 * i^2=2 * 1 * -1=-2

- 1 year, 4 months ago

What is the value of ixi

- 3 years ago

-1

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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

- 2 years, 9 months ago

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

- 1 year, 3 months ago

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

- 1 year ago

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.

- 6 months, 2 weeks ago

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 !!!!

- 6 years, 8 months ago

Yeah it is principal root

- 4 years, 6 months ago

so, what is the modulus and argument?

- 2 years, 6 months ago