Superposition of Plane EM Waves Using Complex Notation

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SUMMARY

The discussion focuses on the superposition of plane electromagnetic (EM) waves using complex notation. The incident electric field is represented as $$\vec{E}_I = E_0 e^{i(ky - wt - \frac{\pi}{2})} \hat{z}$$ and the reflected wave as $$\vec{E}_R = E_0 e^{i(-ky - wt - \frac{\pi}{2})} \hat{z}$$. The summation of these fields leads to the expression $$-2E_0 \cos(ky) \sin(wt) \hat{z}$$, confirming the application of trigonometric identities and complex exponentials in wave analysis. The solution is validated by the community, affirming the correctness of the approach.

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Homework Statement



I have a simple problem relating to the superposition of plane EM waves that I'd to try out using complex notation. Could anyone run through the work to see if my understanding is right?

Many thanks in advance!

The incident E bit of the wave is
$$\vec{E}_I = E_0 \sin(ky - wt) \hat{z} = E_0 \cos(ky - wt - \frac{\pi}{2}) \hat{z}$$
The wave propagating opposite is
$$\vec{E}_R = E_0 \sin(-ky - wt) \hat{z} = E_0 \cos(-ky - wt - \frac{\pi}{2}) \hat{z}$$

I left out the B since it was really the same thing.

Homework Equations

The Attempt at a Solution


First I generalise them to complex notation
$$\vec{\tilde{E}}_I = E_0 e^{i(ky - wt - \frac{\pi}{2})} \hat{z} $$
$$\vec{\tilde{E}}_R = E_0 e^{i(-ky - wt - \frac{\pi}{2})} \hat{z} $$
Then summing them,
$$E_0 \big [ (e^{iky} + e^{-iky})e^{-i(wt + \frac{\pi}{2})} \big] \hat{z} = E_0 \big [ (2\cos(ky))e^{-i(wt + \frac{\pi}{2})} \big] \hat{z}$$
Finally I take the real part
$$E_0 \big [ (2\cos(ky)) \cos(wt + \frac{\pi}{2}) \big] \hat{z} = -2E_0 \cos(ky) \sin(wt) \hat{z}$$

Does this look right?
 
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There is a trig identity that says $$\sin a+\sin b=2\sin \left(\frac{a+b}{2} \right) \cos\left(\frac{a-b}{2} \right).$$ Your expression checks out. Note: This identity can be proven quite readily using complex exponentials, which is what you did.
 
kuruman said:
There is a trig identity that says $$\sin a+\sin b=2\sin \left(\frac{a+b}{2} \right) \cos\left(\frac{a-b}{2} \right).$$ Your expression checks out. Note: This identity can be proven quite readily using complex exponentials, which is what you did.

Thank you!
 

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