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...
Essentially now the question boils down to whether
##Im[Ae^{i(\phi+2p)}+Be^{i(\phi+2p)}+Te^{ip}]=0## is equal to ##Im[A+B+Te^{2ip}]=0## or not, if equal, then what should be the condition on ##\phi## or ##p## I think.
because there is a ##sin(\phi+2p)## being multiplied by ##R##. oh wait. am I correct there? yes you are correct then there should be ##Re^{i(\phi+2p)}## not just ##R##
Okay let me reproduce here, which I got.
Since ##Im[z]=\frac{z-\bar{z}}{2i}## (definition), our ##z## is defined as ##z=RT^*e^{-2ip}## since ##R## is real, so we can write
##\frac{R(T^*e^{-2ip}-Te^{2ip})}{2i}-|T|^2sinp=0##
since ##T## is in general complex, we can replace ##T=re^{i\phi}##, with...
@mfb Thank you. In a physics perspective, if ##p## is a phase, can we choose an overall phase so that ##R## becomes real? Consequently yielding the required equality?
@mfb Thanks. I will recheck the first part of your answer. Regarding the second part I want to clarify that even if ##R## is real, but ##A## and ##B## need not be real, so I kept ##A## and ##B## inside the ##\mathcal{Im}[..]## is it not allowed then?
I have an expression
##\mathcal{Im}[RT^*e^{-2ip}]=|T|^2\sin p ##, where ##R=Ae^{ip}+Be^{-ip} ## and ##p ## is a real number.
This ultimately should lead to ##\mathcal{Im}[A+B+Te^{2ip}]=0 ## upto a sign (perhaps if I didn't do a mistake).
There is a condition on ##R ## that it is real...