How do you find #i^ { - 32} #?
1 Answer
Sep 2, 2017
Explanation:
Note that:
#i^2 = -1#
So:
#i^4 = (i^2)^2 = (-1)^2 = 1#
So:
#i^(-32) = 1/i^32 = 1/((i^4)^8) = 1/(1^8) = 1/1 = 1#
Bonus
The integer powers of
-
#i^(4k) = 1# -
#i^(4k+1) = i# -
#i^(4k+2) = -1# -
#i^(4k+3) = -i#