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How do you simplify #i^64#?

Answer 1

I found: #i^64=1#

I know that: #i^1=sqrt(-1)=i# #i^2=(sqrt(-1))^2=-1# #i^3=i*i^2=i*-1=-i# #i^4=i^2*i^2=-1*-1=1# #i^5=i*i^2*i^2=i*(-1)*(-1)=i# AGAIN #i^6=i^2*i^2*i^2=(-1)(-1)(-1)=-1# now it repeats the same pattern: #i,-1,-i,1# every 4.

You have #i^64=(i^8)^2=(i^4*i^4)^2=(1*1)^2=1#

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Answer 2

James Allen

To simplify i^64, we need to understand the pattern of powers of ( i ). ( i ) raised to any power that is a multiple of 4 simplifies to 1. Since 64 is a multiple of 4 (64 = 4 × 16), ( i^{64} ) simplifies to 1.

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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.