nerdamer i^i returns i instead of e^(-pi/2)

i^i returns i instead of e^(-pi/2)

Open SomebodyLikeEveryBody opened this issue 3 years ago • 0 comments

When we do

nerdamer('i^i').toString()

it returns

instead of

e^(-pi/2)

$i^i=\left(1\cdot e^{\frac{i\pi}{2}}\right)^i=e^{\frac{\pi}{2}\cdot i^2}=e^{-\frac{\pi}{2}}$

Jan 27 '22 23:01 SomebodyLikeEveryBody