One of the most basic thing to keep in mind is that #i^2=-1#. Now if we have any power of #i#, what we need to do is to write it as a product or division of #i^2# and then replace it with #-1#.
Now #i^9=i^(2+2+2+2+1)=i^2xxi^2xxi^2xxi^2xxi#
= #-1xx-1xx-1xx-1xxi=i#
Similarly
#i^15=i^(2++2+2+2+2+2+2+1)#
= #i^2xxi^2xxi^2xxi^2xxi^2xxi^2xxi^2xxi#
= #-1xx-1xx-1xx-1xx-1xx-1xx-1xxi=-i# and
#i^(-4)=1/i^4=1/i^(2+2)=1/(i^2xxi^2)#
= #1/(-1xx-1)=1/1=1#
Also note that #i^1=i#, #i^2=-1#, #i^3=i^2xxi=-1xxi=-i#, #i^4=(i^2)^2=(-1)^2=1#, #i^5=i^4xxi=i#, #i^6=i^4xxi^2=1xx-1=-1#, #i^7=i^4xxi^3=1xx-i=-i# and #i^8=(i^4)^2=(1)^2=1#
Hence successive powers of #i# appear in a cycle of #i#, #-1#, #-i# and #1#. Thus, #i^9=i#, #i^15=-i# and #i^(-4)=1#.
