Question

i^29

Answer 1

i

Answer 2

\[i^{4 \cdot 7+1}=i^{4 c\dot 7} i^1=(i^4)^7i=(1)^7i=1i=i\]

Answer 3

i got it lol thanks to you both!

Answer 4

remainder when you divide 29 by 4 is 1
so you get
\[i^{29}=i^1=i\]

Answer 5

likewise
\[i^{103}=i^3=-i\] and
\[i^{366}=i^2=-1\]