(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?
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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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