How do you simplify #i^109#?

1 Answer
sente
Dec 12, 2015

#i^100 = 1#

Explanation:

We will be using the following:

  • #(x^a)^b = x^(ab)#

  • #i^2 = -1#

Because #i^2 = -1#, we know that #i^4 = (i^2)^2 = (-1)^2 = 1#

Using this, we can take #i# to a large integer power and quickly evaluate it by "pulling out" multiples of #4# from the exponent, like so:

#i^100 = i^4*25 = (i^4)^25 = 1^25 = 1#

While the above suffices for the given problem, it is also applicable to integers which are much larger, or not multiples of #4#. For example, looking at #i^528533#

We can write

#528533 = 528500 + 32 + 1#
# = 100*5285 + 4*8 + 1#
#= 4*(25*5288) + 4*8 + 1#
#= 4(25*5288 + 8) + 1#

So, applying this to our problem:

#i^528533 = i^(1+4(25*5288))#
#= i^1*(i^4)^(25*5288)#
#=i * 1^(25*5288)#
#=i*1#
#= i#

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