Question

How do you simplify `i^-33`?

Answer 1

Algebraically you could do this like this: `i^(-33)=1/(i^33)=1/(i^32*i)=1/((i^2)^16*i)=1/((-1)^16*i)=`
`=1/(1*i)=1/i * i/i =i/(i^2)=i/(-1)=-i`

If you have to use the trigonometric form it goes like this:
`i^(-33)=[1*(cos (pi/2) +i sin (pi/2))]^(-33)=`
`=1^(-33)*(cos((-33pi)/2)+i sin((-33pi)/2))=`
`=cos((33pi)/2)-i sin((33pi)/2)=`
`=cos(8*2pi + pi/2)-i sin(8*2pi + pi/2)=`
`=cos(pi/2)-i sin(pi/2)=`
`=0-i*1=-i`
What I used here is the De Moivre's formula and some basic properties of `sin` and `cos`.