e^iaz

Sep 2009
41
0
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?
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
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?

Then \(\displaystyle az\) must be the multiple of \(\displaystyle \pi\) .
 
  • Like
Reactions: Dgphru

Prove It

MHF Helper
Aug 2008
12,883
4,999
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 \(\displaystyle z = x + i\,y\)...


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

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

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

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

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

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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?
More to the point, where did the "z" go? I suggest you review your notes more carefully since one side involves "z" and the other "x". Do your notes say how those are connected? If not, then this is nonsense.