This is a detailed version of my question posted yesterday. 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, i.e., $R^*=R$, but $A$ and $B$ are not in general real. Further, $T$ depends on $A$ and $B$ in such a way that if $A=0$, $B=0$, then $T=0$, so the (desired) equality holds and $A\neq B$ Here is what I do to achieve the desired result:
$\mathcal{Im}[(Ae^{ip}+Be^{-ip})T^*e^{-2ip}-|T|^2e^{ip}]=0$
Then I take common $T^*e^{-ip}$ from the above expression and it leads me to $\mathcal{Im}[\{(Ae^{2ip}+B)e^{-ip}-Te^{2ip}\}]=0$
This leads to $\mathcal{Im}[Ae^{ip}+Be^{-ip}-Te^{2ip}]=0$
This I rewrite as (since $R=Ae^{ip}+Be^{-ip}$ is real)
$\mathcal{Im}[A+B-Te^{2ip}]=0$, This is the result which is correct upto a sign.
I want to know whether I made a mistake? or there is a mistake in what I want to achieve (regarding the $+$ sign infront of $T$ expression in the desired versus achieved)? Thanks.
Let $\,T=t e^{i \varphi}\,$ where $\,t = |T|\,$ and $\,\varphi \in \Bbb R\,$, then using that $\,R \in \Bbb R\,$:
$$\require{cancel} \begin{align} \operatorname{Im}\left(R \,\overline T \,e^{-2ip}\right)=|T|^2 \sin(p) &\;\;\iff\;\; R\cdot t \cdot \operatorname{Im}\left(e^{-i\varphi}\,e^{-2ip}\right)=t^2 \sin(p) \\ &\;\;\iff\;\; - R \cdot t \cdot \sin(\varphi+2p) = t^2 \sin(p) \end{align} $$
The latter is a relation between the modulus $\,t\,$ and argument $\,\varphi\,$ of $\,T\,$, and represents the necessary and sufficient condition for the proposed equality to hold. For $\,t \ne 0\,$ it reduces to:
$$ |T| \sin(p) + R\,\sin\big(2p+\arg(T)\big) = 0 $$