Question

\[e^{3i\pi/2}\]

Answer 1

\(e^{i 3\pi/2} = \cos(3\pi/2) + i \sin(3\pi/2) \)

Answer 2

=0+i(-1)
=-i

Answer 3

wonder wats wrong wid using euler identity directly:-

\(e^{i3\pi/2} = (-1)^{3/2} = \sqrt{(-1)^3 } = i \) ?

Answer 4

sqrt(-1^3) = sqrt(-1) = i

But (sqrt(-1))^3 = i^3 = -i

????????? whats going on

Answer 5

yahhh somthing fundamental i guess..... @ikram002p have a look if u have time

Answer 6

very fundamental. what am i not seeing?

Answer 7

i think that fundamentally you should take the sqrt(-1) first and call it i.
After which you can go i^3 which is -i

Answer 8

this then agrees with the euler trig expansion