Equating coefficients of complex exponentials

[AtoZ]

I have an equation that looks like

$i\dot{\psi_n}=X~\psi_n+\frac{C~\psi_n+D~a~\psi^\ast_{n+1}+E~b~\psi_{n+1}}{1+\beta~(D~\psi^\ast_{n+1}+E~\psi_{n+1})}$

where $E,b,D,a,C,X$ are constants. I have the ansatz

$\psi_n=A_n~e^{ixt}+B^\ast_n~e^{-itx^\ast}$, $x$ and $A_n,B_n$ are complex. I have to equate coefficients of $e^{ixt}$ and $e^{-itx^\ast}$, I get

$-xA_n~e^{ixt}+x^*B^\ast_n~e^{-itx^*}=\left[X~(A_n~e^{ixt}+B^\ast_n~e^{-itx^\ast})+\frac{C~(A_n~e^{ixt}+B^\ast_n~e^{-itx^\ast})+D~a~(A_{n+1}^*~e^{-itx^*}+B_{n+1}~e^{ixt})+E~b~(A_{n+1}~e^{ixt}+B^*_{n+1}~e^{-itx^*})}{1+\beta~[D~(A_{n+1}^*~e^{-itx^*}+B_{n+1}~e^{ixt})+E~(A_{n+1}~e^{ixt}+B^*_{n+1}~e^{-itx^*})]}\right]$

Now to equate coefficients of say $e^{ixt}$, I get

$-xA_n=XA_n+\frac{C~A_n+D~a~B_{n+1}+E~b~A_{n+1}}{1+\beta(D~B_{n+1}+E~A_{n+1})}$ is true? or the denominator has to be written in full?

 

[mfb]

or the denominator has to be written in full?

Even worse, you'll have to split the fraction properly into one part proportional to $e^{ixt}$ and one proportional to $e^{itx^*}$, if this is possible at all.

 

[AtoZ]

means the one I wrote is incorrect?

 

[mfb]

Yes.

$\frac{a+b}{c+d} \neq \frac a c + \frac b d$