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)=i−33=1i33=1i32⋅i=1(i2)16⋅i=1(−1)16⋅i=
=1/(1*i)=1/i * i/i =i/(i^2)=i/(-1)=-i=11⋅i=1i⋅ii=ii2=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)=i−33=[1⋅(cos(π2)+isin(π2))]−33=
=1^(-33)*(cos((-33pi)/2)+i sin((-33pi)/2))==1−33⋅(cos(−33π2)+isin(−33π2))=
=cos((33pi)/2)-i sin((33pi)/2)==cos(33π2)−isin(33π2)=
=cos(8*2pi + pi/2)-i sin(8*2pi + pi/2)==cos(8⋅2π+π2)−isin(8⋅2π+π2)=
=cos(pi/2)-i sin(pi/2)==cos(π2)−isin(π2)=
=0-i*1=-i=0−i⋅1=−i
What I used here is the De Moivre's formula and some basic properties of sinsin and coscos.
