# Optics: Complex exponentials for sine

1. Jan 31, 2013

### jbrussell93

1. The problem statement, all variables and given/known data
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

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

2. Relevant equations

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

3. 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?

2. Jan 31, 2013

### 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 $\sin\theta$ ...