What is e^(pi*i) using de moivre's?

You will Start with  z=cos(x)+i*sin(x)(1) , = i*z
so dz/z=i*dx   ,   in (z) = i*x+c from (1); when x=0 | z=1 so c=0
                           in (z) = i*x
z=e^(ix) but z= cos (x)+i*sin
so cos (x)+i*sin (x) = e^ (i*x) (x) , Put x=pi in (z) & we get 

ANSWER = -1+0=e^(i*p)  , so it is = e^(i*pi)+1=0

Dose this help you in any way?