How do you simplify #i^333#?

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Answer 1 · 2016-01-13T23:04:24

#i^333=i#

Using the exponent rules:

#a^(n+m)=a^n*a^m#

#a^(n*m)=(a^n)^m#

#i^333=i*i^332=i*(i^2)^(332/2)=i*(i^2)^166#

as #i^2=-1#

#:.i*(i^2)^166=i*(-1)^166#

as #(-1)^(2p)=1#

#:.i*(-1)^166=i*1=i#