(cos theta+i sin theta)^(n)=cos n theta+i sin n theta(i=sqrt(-1))

What is the modulus of the complex number (cos theta+i sin theta)/(cos theta-i sin theta) where i=sqrt(-1) ?

"(cos theta+i sin theta)^(6)(cos theta-i sin theta)^(-3)

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

Prove,by induction,that (cos theta+i sin theta)^(n)=cos n theta+i sin n theta for all positive as well as negative integral values of

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)

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

if theta=(pi)/(6) then the 10 th term of the series 1+(cos theta+i sin theta)^(1)+(cos theta+i sin theta)^(2)+

The value of (1+cos theta+i sin theta)^(n)+(1+cos theta-i sin theta)^n=

The value of (1+cos theta+i sin theta)^(n)+(1+cos theta-i sin theta)^(n)=