How do you simplify #i^1776#?

1 Answer
Dec 18, 2016

1

Explanation:

there is a pattern to powers of #i#:

#i^1 = i#

#i^2 = -1#

#i^3 = -i#

#i^4 = (i^2)^2 = (-1)^2 = 1#

this cycle repeats for every power of #4#.

to simplify #i^1776#, we have to see if #1776# is divisible by #4#, and if not, find the remainder:

#1776/4 = 444#

there is no remainder, meaning that #i^1776# is in the fourth part of the pattern.

this also means that #i^1776# could be simplified as the answer to #i^4#, which is 1.