(or a toddler) Why does “i^i” result in a real number “e^-0.5pi”? How did they figure those out in the first place?
In: 9
It comes from
i = e^(i*pi / 2)
If you raise both sides to the power of i, you get
i^i = e^(i * i * pi/2)
Since i*i is -1, that turns into
i^i = e^(-pi/2)
Proving the original formula is a little more complicated. It comes from Euler’s identity which is
e^ix = cos(x) + i*sin(x)
If you plug pi/2 in for x, cos(x) becomes 0 and sin(x) becomes 1, so it changes to
e^(i*pi/2) = i
As for why Euler’s identity works, I’m not sure I can explain that.
Although this does not have to be very complicated, if one starts from zero, the story would be rather long for a Reddit comment. But there are excellent series of mini-lectures which cover the subject very well. See here: Eddie Woo [“Complex Numbers”](https://www.youtube.com/watch?v=gHUHZXjpwOE&list=PL5KkMZvBpo5CE__2qeqZQa5e8gSkt1Ypy)
> Why does “i^i” result in a real number “e^-0.5pi”?
Technically i^i is multivalued: e^(-pi/2) is just one of the answers, and the only one that is real. You often have to deal with multivalued functions when working with complex numbers (it’s very similar to how 4 has two square roots: 2 and -2).
> How did they figure those out in the first place?
Well, first you have to define what is meant by raising a complex number to the power of another complex number. However, it turns out that there is only one way of doing this such that (a) it reduces to the usual definition of powers when you put real numbers in, and (b) the answers vary smoothly as you change the inputs.
You should know how to think of something like A^(i), when A is postive real, first. Euler’s formula tells us that e^(it) is the point in the Complex Plane which is rotates the number 1 by t radians along the unit circle. Or, without complex numbers or anything, you can just think of e^(it) as just a real fancy way of writing the point on the unit circle corresponding to t radians. So e^(i pi/4) “is” the point (sqrt(2)/2, sqrt(2)/2), for instance.
We can think about it a bit more dynamically too. Imagine you have an arrow rotating around the unit circle with some rotational velocity w. Then the position of the arrow after t-seconds will be e^(iwt). You should just think of e^(stuff) as just a fancy shorthand for rotation.
If we have something like 5^(it), then we can use logs to write it in our fancy shorthand as e^(it ln(5)) and so the “rotational speed” that 5^(it) rotates is ln(5). More generally, the rotational velocity of A^(it) is ln(A).
Something else to clock is that we have another Euler’s formula which says that e^(at) (when a is real) is still a “rotational” thing, it is just with *hyperbolic* rotations. A “hyperbolic rotation” of “hyperbolic angle” x kinda squishes things along hyperbolas rather than rotating around circles. But we can use this to imagine what a *circular rotation* about an *imaginary angle* is. If e^(i wt) rotations around by an angle of wt, then if w=i then the two i’s cancel out to give e^(-t) which is a hyperbolic stretch/squeeze thing. So rotations by “imaginary angles” are just stretches and squeezes.
So i^(it) = e^(it ln(i)) and so the “rotational velocity” is ln(i) which “is” i*pi/2. This means that it will be a hyperbolic rotation rather than a circular one. Which means it will stretch/squeeze things with “velocity” -pi/2 instead. We then get i^(it)=e^(-t pi/2) and, specifically, i^(i) = e^(-pi/2).
(Note: as others mention, there are multiple logs for i so it is actually not unique, but this is the idea)
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