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I have the general recipe for finding the complex conjugate of a function down as follows:

Suppose I have $f(z)$:

  1. Separate $f(z)$ into a sum of real and imaginary functions such that $$f(z)=u(x,y)+iv(x,y)$$
  2. Negate the $i$ and you have your complex conjugate

I have tried the above recipe for the following function(which I obtained from another question on math.SE -- Complex conjugate of function): $$\psi(x,t)=Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)}$$

(I basically didn't understand what they did over there, but I think my procedure is correct, and I tried to rework the problem of finding the complex conjugate using the above process.)

So I write $A=x_1+iy_1$ and $B=x_2+iy_2$, $\theta_1=kx-\omega t$ and $\theta_2=kx+\omega t$, and multiply out the following expression:$$\psi(x,t)=Ae^{i(\theta_1)}+ Be^{-i(\theta_2)}$$using the Euler relation, and I should get the answer, right? I tried this out on pen and paper and I get something like:$$\psi^*=x_1e^{-\theta_1}+x_2e^{\theta_2}-y_1e^{\frac{\pi}{2}-\theta_1}+y_2e^{\frac{\pi}{2}-\theta_1}$$

I am not 100% sure about my result as I could have made a mistake, but I'd like to know if the general procedure is correct. Thanks!

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Hint: $$\forall\,t\in\Bbb R\,\,\,,\,\,e^{it}:=\cos t+i\sin t$$ –  DonAntonio Nov 19 '12 at 11:56
1  
And, using DonAntonio's comment, $\overline{e^{it}}=e^{-it}$. It is also useful to note that $\overline{xy}=\bar x\bar y$ –  Dennis Gulko Nov 19 '12 at 12:29
    
Thanks, I got it. It's much quicker to work with the exponential form. –  Joebevo Nov 19 '12 at 17:00

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