How do you simplify i^-3?

2 Answers
Apr 4, 2016

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

Explanation:

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

Sep 3, 2016

i

Explanation:

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