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

((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

(1)/(cos theta-i sin theta)

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

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

If ((1 + cos theta + i sin theta ) /(i+ sin theta + i cos theta ))= cos ntheta + i sin n theta then n is equal to :

(i) |(cos theta,-sin theta),(sin theta,cos theta)|

[(1 + cos theta + i sin theta) / (sin theta + i (1 + cos theta))] ^ (4) = cos n theta + i sin n theta