`(cos theta+i sin theta)^n = cos ntheta +i sin ntheta`

Using De Moivre 's theorem prove that : ((1+cos theta +i sin theta)/(1+cos theta - i sin theta))^n= cos n theta + i sin n theta , where i = sqrt(-1)

cosm theta - sin ntheta =0

(cos theta + i sin theta) ^ (n) = cos n theta + i sin n theta

Prove that ((1+cos theta+i sin theta)/(1+cos theta-i sin theta))^(n)=cos n theta+i sin n theta

(cos theta + i sin theta) (cos3 theta + i sin3 theta) .......... (cos (2r-1) theta + i sin (2r-1) theta) = 1

The number of values of theta in (0, pi] , such that (cos theta + i sin theta) (cos 3 theta + i sin 3 theta) (cos 5 theta + i sin 5 theta) (cos 7 theta + i sin 7 theta) (cos 9 theta + i sin 9 theta) = -1 is

(cos theta + i sin theta)^3 xx (cos theta-i sin theta)^4=

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