Question 1207629


1.

{{{2i(i^(-3))}}}

={{{2i/i^3}}}
={{{2/i^2}}}...........since {{{i^2=-1}}}
={{{2/-1}}}
={{{-2}}}

 in the form {{{a + bi}}} {{{a=-2}}} and {{{b=0}}}, so you have 

{{{-2}}}



2. {{{2i(i^(-3))/i^(-4)}}}....from 1. we have that {{{2i(i^(-3))=-2}}}

={{{-2/i^(-4)}}}.........{{{i^(-4)=1/i^4=1/(i^2)^2=1/(-1)^2=1}}}

={{{-2/1}}}

={{{-2}}}