e^iaz

Dgphru

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
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 $$\displaystyle \pi$$ .

Dgphru

Prove It

MHF Helper
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
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.