Question

Which of the following is equivalent to the complex number i^-9

Answer 1

the solution is

$-i$

because we know

$i^{-2}=-1$

and

$i^{-1}=-i$

so

$\begin{gathered} i^{-2}\times i^{-2}\times i^{-2}\times i^{-2}\times i^{-1}=i^{-9} \\ -1\times-1\times-1\times-1\times \end{gathered}$