Question

question in comments

Answer 1

\[e^{-i n (\pi/2)} = (-i)^{n}\]

Answer 2

how to show that this equation is true?

Answer 3

\[e^{i\theta}=\cos(\theta)+i\sin(\theta)\]

\[e^{-in(\pi/2)}=\left(e^{i(-\pi/2)}\right)^n\]

\[=\left(\cos(-\pi/2)+i\sin(-\pi/2)\right)^n\]

\[=\left(0+i(-1)\right)^n=(-i)^n\]