How do you simplify #3i^42 - 5i^46 + 2i^29 - (4i^2)^3 + i^0#?
1 Answer
Dec 6, 2015
#3i^42-5i^46+2i^29-(4i^2)^3+i^0 = 67+2i#
Explanation:
For any integer
#i^(4k+0) = 1#
#i^(4k+1) = i#
#i^(4k+2) = -1#
#i^(4k+3) = -i#
So:
#3i^42-5i^46+2i^29-(4i^2)^3+i^0#
#=3(-1)-5(-1)+2(i)-4^3(-1)+1#
#=-3+5+2i+64+1 = 67+2i#
