# Verifying an equality

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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 ##, and ##A\neq B## so the (desired) equality holds. 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 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 plus sign in front of $T$ expression in the desired versus achieved)? Thanks.

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Thanks, I updated the post with Latex now.

##\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 ##
Something went wrong with the exponentials here. In addition you can't just divide by a complex number, the imaginary part might change. As a simple example, ##Im(1)=0## but ##Im(\frac{1}{i}) \neq 0##.

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

##\mathcal{Im}[A+B-Te^{2ip}]=0 ##
That doesn't look like a correct mathematical operation. If ##R=Ae^{ip}+Be^{-ip} ## is real and you take the imaginary part of your expression then you should just remove it from the sum, without leaving in A and B.

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Something went wrong with the exponentials here. In addition you can't just divide by a complex number, the imaginary part might change. As a simple example, ##Im(1)=0## but ##Im(\frac{1}{i}) \neq 0##.
That doesn't look like a correct mathematical operation. If ##R=Ae^{ip}+Be^{-ip} ## is real and you take the imaginary part of your expression then you should just remove it from the sum, without leaving in A and B.

@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?

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?
You can't just randomly decide to "keep something in" in some modified version. As an example, consider ##A=B=i##, ##p=\frac \pi 2##. Then ##R=Ae^{ip}+Be^{-ip} = i^2-i^2 = 0## but ##\mathcal{Im}(A+B)=2##.

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You can't just randomly decide to "keep something in" in some modified version. As an example, consider ##A=B=i##, ##p=\frac \pi 2##. Then ##R=Ae^{ip}+Be^{-ip} = i^2-i^2 = 0## but ##\mathcal{Im}(A+B)=2##.
@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?

I thought p was some unknown constant. Since when can we choose it? For given A,B there will always be value of p where R is real. For p=0 and p=pi you get opposite imaginary parts and the imaginary part is continuous in p, therefore there must be a zero crossing.

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I thought p was some unknown constant. Since when can we choose it? For given A,B there will always be value of p where R is real. For p=0 and p=pi you get opposite imaginary parts and the imaginary part is continuous in p, therefore there must be a zero crossing.
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 arbitrary ##\phi## and that ##|T|=r##
##-\frac{R|T|\left[e^{i(\phi+2p)}-e^{-i(\phi+2p)}\right]}{2i}-|T|^2sinp=0## becase ##r=|T|##,
##Rsin(\phi+2p)+|T|sinp=0##
which can we rewrite as?
##Im[Re^{i(\phi+2p)}+Te^{ip}]=0##, now here the assumption of overall phase could matter that "we choose an overall phase so that $R$ is real, and that leads to the expression which is desired i.e., ##Im[A+B+Te^{2ip}]=0##.

One more thing: We can make ##R## real only in the following case
if we make the replacement ##A=c_1e^{-ip}## and ##B=c_2e^{ip}## then ##R## is real. which essentially means that ##R=c_1+c_2##, but we can rename ##c_1## and ##c_2## as ##A## and ##B## later to match the result with the required equality.
So choosing and overall phase so that ##R## is real, makes sense with this argument? My intuition regarding these phase and stuff is bad, sorry.

I suppose I am quite close but still not exactly there.

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which can we rewrite as?
How?

If R is real, then ##Im[R+Te^{ip}]=Im[Te^{ip}]##.

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How?

If R is real, then ##Im[R+Te^{ip}]=Im[Te^{ip}]##.
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##

That is not an explanation how you got the following expression. Can you break it down step by step?

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That is not an explanation how you got the following expression. Can you break it down step by step?
Okay, I have modified my above answer with more steps. Can you please see whether I made a mistake?

That is not an explanation how you got the following expression. Can you break it down step by step?
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.

They are not equal in general, see my example a few posts ago.

They are identical if ϕ+2p is a multiple of 2pi (trivial). They can be identical for other values, that depends on A and B.
Okay, I have modified my above answer with more steps.