wat is i to the power i?

Question

i - root of -1

Answer 1

i^i = e^(-π/2)

proof:

x = i^i

ln(x) = iln(i)
ln(x) = iln(-1^1/2)
ln(x) = (i/2)ln(-1) {since ln(-1) = iπ }
ln(x) = -π/2

thus x = e^(-π/2)

Answer 2

I agree with the answer above.

Answer 3

e^(ix) = cos(x) + i.sin(x)
log(cos(x) + i.sin(x)) = ix

cos(x) + i.sin(x) = i
cos(x) = 0
sin(x) = 1
x = pi/2

log(i) = (pi/2)i

log(i^i) = i*log(i) = (pi/2) * i * i = -pi/2
i^i = e^(-pi/2) = 0.2079 (approx)

Answer 4

0.207879576