`((cos theta+i sin theta)^(4))/((sin theta)+i cos theta)^(2)`

((cos theta+i sin theta)^(4))/((sin theta+i cos theta)^(5)) is equal to.

((cos theta-i sin theta)^(4))/((sin theta+i cos theta)^(5)) is equal to

((cos theta+i sin theta)^(4))/(((sin theta+i cos theta))^(5)) is equal to

((cos theta + i sin theta) / (sin theta + i cos theta)) ^ (4) = cos8 theta + i sin8 theta

In the form of A+iBz=(cos2 theta+i sin2 theta)^(-5)(cos3 theta-i sin3 theta)^(6)(sin theta-i cos theta)^(3) is

(((cos theta+sin theta)^(4)-(cos theta-sin theta)^(4))/(sin theta cos theta))=?

((cos2 theta+i sin2 theta)^(3)(cos3 theta-i sin3 theta)^(5))/((cos3 theta+i sin3 theta)^(2)(cos4 theta+i sin4 theta)^(3))=

((cos3 theta+i sin3 theta)^(5)(cos theta-i sin theta)^(3))/((cos5 theta+i sin5 theta)^(7)(cos2 theta-i sin2 theta)^(5))