How do you simplify #i^13#?

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Answer 1 · 2016-10-21T18:01:44

#i#

#i=sqrt-1#

#i^2=-1#

#i^3=i#

#i^4=1#

multiplying bi #i# repeats the pattern.

So any #i^(4n)=i#

Since #13-=1(mod4)#

#i^(13)=i^1=i#