show that
exp (i (x + y)) = exp (ix) exp (iy)
thanks for your reply. i ended up with this :
exp(i(x+y))= cos(x+y)+isin(x+y)
= cos(x)+cos(y)+isin(x)+isin(y)
=[cos(x)+isin(x)]+[cos(y)+isin(y)]
where cos(x)+isin(x)= exp(i(x))
after substitution we get
=exp(i(x))+exp(i(y))
the confusing part is the "+" sign that i'm getting it should be " * "
any suggestions..
please take a look at this
The Way of Analysis - Robert S. Strichartz - Google Books
I'm using the same method but my answer is ... exp(i(x))+exp(i(y)) which is not correct
thank you for your help i got it
cos(x+y)=cosx.cosy-sinx.siny
sin(x+y)=sinx.cosy+cosx.siny
Now, cos(x+y)+isin(x+y)
=(cosx.siny-sinx.cosy)+i(sinx.cosy+cosx.sinx)
= cosx.siny-sinx.cosy+isinx.cosy+icosx.sinx
=(cosx+isinx)(cosy+isiny)[i^2=-1]