How do you find #i^ { - 32} #?

1 Answer
Sep 2, 2017

#i^(-32) = 1#

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#

#color(white)()#
Bonus

The integer powers of #i# cycle round every #4# powers, so we find that for any integer #k#:

  • #i^(4k) = 1#

  • #i^(4k+1) = i#

  • #i^(4k+2) = -1#

  • #i^(4k+3) = -i#