How do you simplify #i^109#?

Answer 1

#i^100 = 1#

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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