Earlier quoted context omitted.
Your intuition is nice for figuring out the directions, but not so nice for explaining the numbers. By which I mean, I can understand why it's real, but I can't explain why i^i = e^(-pi/2)... I would expect it to be a nicer number instead (like 1/2, or 1/e, or something like that). I can't really explain why both pi and e end up int he formula.
Oh yeah, after getting the direction, figuring out the numbers is the next step :). Having i as a base means "we plan on rotating 90 degrees" which actually means pi/2 radians. e^rt models growth rate of r, for time of t. so e^(i · pi/2) creates a 90 degree turn (we intend on rotating, i, and do this enough to get a full 90-degree turn, pi/2). This is all a fancy way of saying: 90 degree turn = i = e^(i · pi/2) Now,…
1. Here's a deeper intuition: any circle is just the unit circle, scaled up or down. Any number is just 1.0, scaled (and rotated, if complex) by the exponential function that was run for some rate and for some amount of time. e^rt is a rocket ship of constant change, we just decide how long to stay on for, and we can get to any number.
In other words, for any number a: a = 1.0 * e^ln(a)
This formulation is useful if we know we're going to be taking exponents on our number a, i.e. we really want a^b. (If we know we'll be rotating our number, maybe we write it in polar coordinates, etc.)
So, the intuition is: "I know I'm going to be taking my number to various powers, so let's get the base settings for e^rt dialed in. pi/2 is the setting for how long we'd ride e^i for in order to get to 90 degrees. I should expect pi/2 somewhere in the answer as I take it to various powers."
2) I haven't taken complex analysis, so my understanding isn't nuanced enough here either. Technically, i^i can be multi-valued, for this graphical analogy let's settle on the principal root (https://www.math.hmc.edu/funfacts/ffiles/20013.3.shtml).
> (e^(2 i pi))^(1/2) is not e^(i pi), it's e^0.
(e^(2 i pi))^(1/2) is asking for the square root of 1, which is both 1 [e^0] and -1 [e^(i pi)]. Again there may be a subtlety here, but I'm not sure how the above statement is incorrect (barring a technicality like "we always mean the positive root"). For the purposes of an intuition it makes sense, I think.