How do you simplify i^64?

1 Answer
Nov 15, 2015

I found: i^64=1

Explanation:

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