Vtyjkyuk

This is a detailed version of my question posted yesterday. I have an expression
$mathcalIm[RT^*e^-2ip]=|T|^2$sin $p$, where $R=Ae^ip+Be^-ip$ and $p$ is a real number.

This ultimately should lead to $mathcalIm[A+B+Te^2ip]=0$ 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 $Aneq B$ Here is what I do to achieve the desired result:

$mathcalIm[(Ae^ip+Be^-ip)T^*e^-2ip-|T|^2e^ip]=0$

Then I take common $T^*e^-ip$ from the above expression and it leads me to
$mathcalIm[(Ae^2ip+B)e^-ip-Te^2ip]=0$

This leads to $mathcalIm[Ae^ip+Be^-ip-Te^2ip]=0$

This I rewrite as (since $R=Ae^ip+Be^-ip$ is real)

$mathcalIm[A+B-Te^2ip]=0$, 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 $,R in Bbb R,$:

$$requirecancel
beginalign
operatornameImleft(R ,overline T ,e^-2ipright)=|T|^2 sin(p)
&;;iff;; Rcdot t cdot operatornameImleft(e^-ivarphi,e^-2ipright)=t^2 sin(p) \
&;;iff;; - R cdot t cdot sin(varphi+2p) = t^2 sin(p)
endalign
$$

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:

$$
|T| sin(p) + R,sinbig(2p+arg(T)big) = 0
$$

Your conclusion is not correct (either way). For a counterexample, take $A=B=1$, $T=4(cos p)^2$ and $0<p<pi/4$ arbitrary. Note that $R=2cos p$ in this case.