How do you simplify # i^-3#?

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Answer 1 · 2016-04-04T05:17:24

#i^-3 =i= sqrt(-1)#

this is the correct explanation.

#i^2 =(-1)#

#i = sqrt(-1)#

#a^-n = 1/a^n#

#i^-3 = (1/i)^3#

#=1/sqrt(-1)^3#

#=1/(sqrt(-1)*sqrt(-1)*sqrt(-1))#

#=1/(-1*sqrt(-1))#

#=-1/sqrt(-1)#

Rationatise the denominator by multiplying the numerator and denominator by #sqrt(-1)#

#=(-1*sqrt(-1))/(sqrt(-1)*sqrt(-1))#

#=-sqrt(-1)/ -1#

#=sqrt(-1)#

#=i#

Answer 2 · 2016-09-03T17:36:37

#i#

First, note that:

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

Also, using the rule #a^-b=1/a^b#, we see that:

#i^-3=1/i^3#

Multiply it by #i/i#:

#1/i^3*i/i=i/i^4=i/1=i#