Question

What is `(1+i)^i` ?

Answer 1

`(1+i)^i = e^(- (2npi + pi/4)) (cos(1/2 ln(2)) + i sin(1/2 ln(2)))`

for any integer `n`, with principal value when `n=0`

Explanation:

Note that:

`1+i = sqrt(2) e^((2npi + pi/4) i) = e^(1/2 ln(2) + (2npi + pi/4) i)" "` for any integer `n`

So:

`(1+i)^i = (e^(1/2 ln(2) + (2npi + pi/4) i ))^i`

`color(white)((1+i)^i) = e^(1/2 ln(2)i - (2npi + pi/4))`

`color(white)((1+i)^i) = e^(- (2npi + pi/4)) (cos(1/2 ln(2)) + i sin(1/2 ln(2)))`

for any integer `n`, with principal value when `n=0`.