Question

Sinx+iCosx/1+i=0 solve it

Answer 1

Answer:

Using the de Moivre's identity which states

eix=cosx+isinx we have

1+eix1+e−ix=eix1+e−ix1+e−ix=eix

NOTE

eix(1+e−ix)=(cosx+isinx)(1+cosx−isinx)=cosx+cos2x+isinx+sin2x=1+cosx+isinx

or

1+cosx+isinx=(cosx+isinx)(1+cosx−isinx)