How do you simplify #i^123458#?

1 Answer

Factor out #i^4#

Explanation:

The nice thing about powers of "i" is that there are only four possible answers: i, -i, 1, and -1.
Since #i=sqrt(-1)# then #i^4=(i^2)^2=1#. "i" to a power of any multiple of four ( #i^4,i^8,i^16,...)# is 1 , so factor out i to the power that is the next lower multiple of four, in this case 123456.

#i^123458=i^123456*i^2=1*i^2=(sqrt(-1))^2=-1#