Optics: Complex exponentials for sine

[jbrussell93]

Homework Statement

Imagine that we have two waves of the same amplitude, speed and frequency over-lapping in some region of space such that the resultant disturbance is

\psi(y,t) = Acos(ky+\omega t) + Acos(ky-\omega t +\pi)

Using complex exponentials show that


\psi(y,t) = -2Asin(ky) sin(\omega t)

Homework Equations

Ae^{i\theta} = Acos(\theta) + Aisin(\theta)

The Attempt at a Solution

I actually have the solution worked out, but my issue occurred once consulting the solutions manual. I ended up with:

\psi(y,t) = A(2icos(ky)sin(\omega t)-2sin(ky) sin(\omega t))

with the real part being the solution:

\psi(y,t) = -2Asin(ky) sin(\omega t)



Then the solutions manual says "Had we begun with sine waves, i.e.

\psi(y,t) = Asin(ky+\omega t) + Asin(ky-\omega t +\pi)

the treatment would have been identical up until the last step, where, this time, the imaginary part would be taken to give:"

\psi(y,t) = 2Acos(ky)sin(\omega t)


I'm confused as to why we can just choose the imaginary part because we happen to have started with sine functions. I thought the real part was always used and the imaginary part ignored. I think I'm missing some intuition with converting sine functions into complex exponentials... I understand that

Acos(\theta) = Ae^{i\theta} because we can ignore the Aisin(\theta) term,
but what if we start with Asin(\theta)? How is this changed to an exponential?

 

[Simon Bridge]

You don't. The cosine-sine part is the real part when you start with sine functions.
This is because:

$\sin\theta = \frac{1}{i}\left ( e^{i\theta}-e^{-i\theta} \right )$

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LaTeX note: put a backslash in front of function names to get them to typeset properly ... sin\theta gives you $sin\theta$ while \sin\theta gives you $\sin\theta$ ...