Yes, that is true, of course. But consulting a Lisp REPL merely demonstrates that it is true. It does not explain why it is true.
You'd get further, pedagogically speaking, by pointing out that (e^i)^i = e^(i*i) = e^(-1) = 1/e, and so e^i is a number that is in some sense "half way" between e and 1/e when you are exponentiating. As an analogy, consider:
e^(1/2) * e^(1/2) = e^(1/2 + 1/2) = e^1 = e
as a demonstration that e^(1/2) is a number that is in some sense "half-way" between 1 and e when you are multiplying, a.k.a. the square root of e. But that still leaves unanswered the question of what "half-way" means when exponentiating rather than multiplying.
Comments
Yes, that is true, of course. But consulting a Lisp REPL merely demonstrates that it is true. It does not explain why it is true.
You'd get further, pedagogically speaking, by pointing out that (e^i)^i = e^(i*i) = e^(-1) = 1/e, and so e^i is a number that is in some sense "half way" between e and 1/e when you are exponentiating. As an analogy, consider:
e^(1/2) * e^(1/2) = e^(1/2 + 1/2) = e^1 = e
as a demonstration that e^(1/2) is a number that is in some sense "half-way" between 1 and e when you are multiplying, a.k.a. the square root of e. But that still leaves unanswered the question of what "half-way" means when exponentiating rather than multiplying.
Obtaining an intuition for what it means to use an imaginary exponent takes work.
Exponents give us expoential decay/growth in the real number line, but periodicity in the imaginary domain, which is strange.
In physics, this lets us analyze decaying or amplifying oscillations in a unified way. (Laplace transform and all that.)