soo this is another stupid question, but i thought that

\(\displaystyle e^{iaz} \)

would equal

\(\displaystyle cos az + i sin az \)

but in my lecture notes it says

\(\displaystyle e^{iaz} = cos ax \)

i dont get it where did the sin go?

I assume \(\displaystyle z = x + i\,y\)...

\(\displaystyle e^{i\,a\,z} = e^{i\,a\,(x + i\,y)}\)

\(\displaystyle = e^{-a\,y + i\,a\,x}\)

\(\displaystyle = e^{-a\,y}e^{i\,a\,x}\)

\(\displaystyle = e^{-a\,y}[\cos{(a\,x)} + i\sin{(a\,x)}]\)

\(\displaystyle = e^{-a\,y}\cos{(a\,x)} + i\,e^{-a\,y}\sin{(a\,x)}\).