# Verifying an equality

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• AtoZ

#### AtoZ

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.

AtoZ
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##.

AtoZ
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}]##.

AtoZ
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?

AtoZ
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.