Question

what is i^-i

Answer 1

plz explain all the steps

Answer 2

i can be written as
\( \cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right) \)

or in polar form as
\( e^{i\frac{\pi}{2} } \)

raise it to the -i th power:
\( \left(e^{i\frac{\pi}{2} } \right)^{-i}\)

use the rule \( \left(a^b\right)^c = a^{bc} \) to simplify

Answer 3

plz...further simplify it /// @phi

Answer 4

are you asking what is -i * i * pi/2 ?

Answer 5

yup......i have tried to solve it i m getting the wrong ans

Answer 6

what is -i * i = - ( i * i) ?
remember i = sqr(-1)