What is i^47?

Powers of $i$ repeat every four steps: $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$. Divide the exponent by 4 and use the remainder: $47 = 4(11)+3$, so $i^{47} = i^3 = -i$, choice J. Choice G ($i$) results from using a remainder of 1 instead of 3. Choice H ($-1$) is what a remainder of 2 would give, and choice F ($1$) is the value when 47 is a multiple of 4, which it is not (44 and 48 are). Always check the remainder carefully, since each possible remainder of 0, 1, 2, or 3 gives a different answer.