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What does i^i = ?

What is i^i?
Before we look at that, I want to have a quick look at Euler's identity.
And everyone gets very, very excited about
e^iπ (or πi which ever way around you want to do it) + 1 = 0
Of course, we can rearrange that a little bit. e^iπ just equals -1.
And Euler's identity is amazing. Don't get me wrong.
I think it's incredible, but... I believe we can do a lot better. In fact, even Euler's identity,
I tend to think of it not like this is one specific case, but as Euler's formula.
So this is the generalized form, you've got
e^iθ = cos(θ) + i sin(θ)
And this is arguably more beautiful than just that one particular case, because what you have here
are the two major ways of representing complex numbers
The e^iθ that θ is just the angle around.
It's the polar coordinate version of a complex number.
And you can scale that in and out to get everywhere on the complex plane.
And then underneath here the cos(θ) + i sin(θ),
that's just the Cartesian coordinates.
So I like this because it's converting between the polar and the Cartesian versions of complex numbers.
We can now put some values in for θ and we get some fantastic identities
If our angle is π/2, e^iπ/2 equals...
i
Pretty straightforward!
Here's the famous one if we go through a full 180 degrees...
e^iπ = -1
But we can carry on
e^i(3π/2) = -i
That's kind of fun. And finally e^i2π just equals 1
Although you could argue I shouldn't just use 2π for that, should I?
I mean really because it's also the angle 0.
So, there are, there are two values going on here.
And I guess the other issue is because θ is an angle, you loop back on yourself,
so you could carry on e^i(5π/2)
also equals i, we've got we've got two versions of that, and that's fine.
So here we have all the fantastic identities, including the particularly, uh, famous one over there,
but it's just examples of Euler's formula.
What I want to look at is... the much more exciting calculation of what is i... to the power of i.
What is the imaginary number to the power of the imaginary number?
Is it going to be complex? Is it going to be real?
Is it just in terms of i? Is it a whole new type of number?
Again, so to work it out, our big issue here is prying apart the i in the base from the i in the power.
I'm going to use a little trick to do that.
I'm going to do a, uh, kind of a function on i^i, and then I'm gonna do the inverse of that.
So, for example I could... square it, and then take the square root. And we're back where we started.
Or I could take the inverse of the inverse, or I could take e to the power of the natural log,
and in theory we'll still get the value of i^i at the other end.
There's a few dangers, because things like... if we square it and take the square root,
if i^i happens to be a negative value, then we'll lose that.
So, there are some nuances to keep an eye on, but for the most part, this is a pretty powerful technique.
We're going to actually use e to the natural log of i^i, and we're doing that because...
...with the logarithm, if you've got something to the power of something inside it,
you can take the exponent, you get the power, and pop it out the front.
And that's how we're going to crack the two i's apart. We now have to work out... what is the natural log of i.
Well we saw that a moment ago, when we were doing our ridiculous things, uh, on the complex, uh, plane.
Sorry, just the screens... get a little crowded here.
And, uh, we can just read off from before that e^i(π/2) equals i,
and the natural logarithm is just whatever you have to raise e to to get the value.
So to get i, e is to the power of... i(π/2).
So we take that and we pop it over... in there. So the natural logarithm...
we've got two i's next to each other. The imaginary number times the imaginary number is -1 by definition.
We swap that out for a negative! And there you have it.
i^i = e^(-π/2)
How cool is that?! I mean, I think that is an incredible result.
The imaginary number to the imaginary number is e^(-π/2).
Now, what's partly so exciting about that, is everything on the right-hand side over there, is a real number.
So we could- we can slap some values in there...
And it turns out that i^i equals 0.207879... and a bunch of other digits after that.
And it's about a fifth.
That's the moral of this video. i^i is about 1/5th.
The other fascinating thing about this...
...is yes, people go on and on about Euler's identity because it's got negative values in there.
It's got e, it's got π, it's got the imaginary number, and it's got powers.
But to my mind, i^i has all of those things.
It's got powers, e, π, i, it's also got... it's got a fraction.
That depends on how you define your circle constant, to be honest.
We should have a value of π, which is half the normal value.
How great would that b- that would just solve a lot of arguments,
and, you know negative numbers... It's all there.
What I like even more is on the left... is completely imaginary. i^i. No real numbers.
On the right is completely real, and people think that, you know, imaginary numbers are kind of an add-on.
A bonus... exotic type of number.
No! They're- they're perfectly valid numbers. The word imaginary... is not a great name, if I'm being entirely honest.
And people, you know, you're aware that if you square i, you get a completely real answer,
so people are vaguely aware there's a link between the imaginary and the real numbers.
But here!
Nothing but imaginary, gives you nothing but real. They are all the same type of numbers.
So there you are, that's my video about i^i.
Thank you so much for watching. Thank everyone who subscribes to my-
Ah, kidding.
Of course I'm going to talk about the bonus values. So people are already yelling at their screens.
I didn't point out, because of the way I did the natural log of i, it's got more than one valid result.
As well as π/2, I could have used 5π/2. I could have used 9π/2.
There are loads. Just loads of, you know, infinitely many some would say, options.
And, uh, the solution here, when it comes to this little problem, is...
Don't worry about it. Just, just, just, just- it's about a FIFTH.
So there you are. Thanks for watching the video.
If you want to learn more about Euler's identity, then 3Blue1Brown has some fantastic videos.
Both, the very first video that was on the channel, and the one that was done two years afterwards.
Please do check those out if you want to learn more about my...
second-favorite, ridiculous identity, involving these constants.