Just Because Today is π Day...

@Lanyx Euler's identity! Yeaahhh! 🥳

@helenS Exactly, except I may have made mistake in it: The -1 should be +1:
e^(i.pi) + 1 = 0, since e^(i.pi) = -1

@Lanyx You're the guy with the Python CMATH module, right?

@helenS I think yes. I do have python, but I haven't played with it in a while.

@Lanyx If I remember correctly, you confirmed that i^(-i) is a (real!) number between 0 and 10.

@helenS I do remember doing (i)^(+i). Confirming it to be around 0.2; So, i^-i should be around 5

@Lanyx i^(-i) is equal to exp(π/2)

@helenS You are evil...
You made me confirm and get stuck. It turns out that python cmath library does not support exponentiation of two complex numbers.
I had to go into my library and find a code which I wrote earlier to do this operation.
I got an answer of around 4.8. Tracing it back, I confirmed your e^(pi/2)

@Lanyx It's easy to prove that the result is exp(π/2).
You just have to use [i]i=exp(i*π/2)[/i], so i^(-i) = (exp(i*π/2))^(-i) where –i=1/i, of course, so (exp(i*π/2))^(-i) = exp(-i/i*π/2) = exp(π/2).
Evil Helen 😈

@helenS I know about that trick. A while back, I wrote a python routine which does (a + bi)^(c + di). I did some manipulation and it spits out an answer.