What does #i^4# equal?

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Answer 1 · 2017-02-01T13:36:40

#i^4=1#

By definition, #i=sqrt(-1)#.

#i^4=sqrt(-1)^4#

When we have a number, say #sqrt2# and we multiply it by another #sqrt2#, we get what's inside the square root sign:

#sqrt2xxsqrt2=2#

So let's apply that to our problem:

#sqrt(-1)^2=-1#

But our problem has four #i#, not two. So let's break it down into two sets of two:

#sqrt(-1)^4=sqrt(-1)^(2+2)=sqrt(-1)^2xxsqrt(-1)^2=-1xx-1=1#

We could also have done it this way:

#sqrt(-1)^4=sqrt(-1)^(2xx2)=(sqrt(-1)^2)^2=(-1)^2=1#