Complex conjugate of function

I have a wavefunction $\psi(x,t)=Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)}$. $A$ and $B$ are complex constants.

I am trying to find the probability density, so I need to find the product of $\psi$ with it's complex conjugate. The problem is, im not sure what is it's complex conjugate, I know the complex conjugate of $5+4i$ is $5-4i$, but what would be the complex conjugate of $\psi$? Is it just $-Ae^{i(kx-\omega t)}-Be^{-i(kx+\omega t)}$?

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The complex conjugation factors through sums and products. So you can take the complex conjugate of the factor with A and B separately. The constant A and B form know problem, this goes according to the usual rules. This leaves something of the form $e^{(a+bi)}$. Now note that $e^{(a+bi)}= e^a(\cos(b)+i \sin(b))$ Taking the complex conjugate now and using $\cos(b)=-\cos(b)$ and $-\sin(b)=\sin(-b)$, you find the complex conjugate $e^{a+i(-b)}$.

This means: $\bar{\psi} = \bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)}$

Note the use of the minus sign to compactly write the complex conjugate of $e^{a+ib}$. In computation this is what write, but you might want to keep the explanation of the $\cos$ and $\sin$ in the back of your head.

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It's easier to note once and for all that $\overline{e^z}=e^{\overline z}$, because the exponential function can be defined uniquely without committing to a choice between $i$ and $-i$ (e.g., by power series). –  Henning Makholm Jan 1 '12 at 18:42
Thanks, so I want to find $\psi\bar{\psi}$. Do I just multiply it out like this? $(Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)})(\bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)})$, and so i get: $A\bar A+B\bar B+A\bar Be^{i(kx-\omega t)+i(kx+\omega t)}+B\bar Ae^{-i(kx+\omega t)-i(kx-\omega t)}$? Is that correct? –  Thomas Jan 1 '12 at 18:46
Yes, if needed you can further simply to: $A\bar A+B\bar B+A\bar Be^{2ikx}+B\bar Ae^{-2ikx}$ –  MrOperator Jan 1 '12 at 18:52

The complex conjugation will map $A \to \bar{A}$ and $B \to \bar{B}$.

If, say $A= 5 + 4 i$, then $\bar{A} = 5 - 4 i$, as you noted. So $$\bar{\psi} = \bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)}$$

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