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I've been trying to solve this question for a while, tried different approaches, but I can't get to the answer. The question is:

Define $\, \, \,e^{i\beta \Phi}=\rho \left ( \dfrac{\tau_0}{\tau_1} \right )^2$

with $\Phi = \Phi_1 + i\Phi_2$, and being that $\Phi_1, \Phi_2, \rho, \tau_0, \tau_1$ are complex numbers and $\beta$ is a real parameter. Show that:

$\Phi_1 = \dfrac{1}{\beta} \tan^{-1} \left [ i \dfrac{e^{-i\delta}\tau_{0}^{*} \tau_1 - e^{i\delta}\tau_0 \tau_{1}^*}{e^{i\delta}\tau_0 \tau_{1}^*+e^{-i\delta}\tau_{0}^{*} \tau_1} \right ]\,\,\,\,\,\,\,\,$ and $\,\,\,\,\,\,\,\, \Phi_2=-\dfrac{1}{\beta}\log \left [ \left | \rho \right | \dfrac{\tau_0^* \tau_0}{\tau_1^* \tau_1} \right ]$, with $\rho = \left | \rho \right |e^{i\delta}$

The following part is the best I could do:

First, I worked out the expression

$e^{i\beta(\Phi_1 + i\Phi_2)}=\left | \rho \right |e^{i\delta}\left ( \dfrac{\tau_0}{\tau_1} \right )^2\qquad$ (1)

Took the conjugate

$e^{-i\beta(\Phi_1^* - i\Phi_2^*)}=\left | \rho \right |e^{-i\delta}\left ( \dfrac{\tau_0^*}{\tau_1^*} \right )^2\qquad$ (2)

Rewrote like this

$e^{i\beta(\Phi_1^* - i\Phi_2^*)}=\dfrac{1}{\left | \rho \right |}e^{i\delta}\left ( \dfrac{\tau_1^*}{\tau_0^*} \right )^2\qquad$ (3)

Then I multiplied (1) and (3) and got the following expression

$e^{i\beta (\Phi_1 + \Phi_1^* + i(\Phi_2 - \Phi_2^*))}=e^{2i\delta}\left(\dfrac{\tau_0 \tau_1^*}{\tau_0^* \tau_1}\right)^2$

$e^{2i\beta (Re(\Phi_1) - Im(\Phi_2))}=e^{2i\delta}\left(\dfrac{\tau_0 \tau_1^*}{\tau_0^* \tau_1}\right)^{2}$

$e^{i\beta (Re(\Phi_1) - Im(\Phi_2))}=e^{i\delta}\left(\dfrac{\tau_0 \tau_1^*}{\tau_0^* \tau_1}\right)\qquad$ (4)

Also, multiplying (1) and (2), I got:

$e^{i\beta (\Phi_1 - \Phi_1^* + i(\Phi_2 + \Phi_2^*))}= \left | \rho \right |^{2}\left(\dfrac{\tau_0 \tau_0^*}{\tau_1 \tau_1^*}\right)^2$

$e^{i\beta (2i Im(\Phi_1) + 2i Re(\Phi_2))}= \left | \rho \right |^{2}\left(\dfrac{\tau_0 \tau_0^*}{\tau_1 \tau_1^*}\right)^2$

$e^{-\beta (Im(\Phi_1) + Re(\Phi_2))}= \left | \rho \right |\left(\dfrac{\tau_0 \tau_0^*}{\tau_1 \tau_1^*}\right)\qquad$ (5)

(5) gives me directly that:

$Im(\Phi_1) + Re(\Phi_2) = -\dfrac{1}{\beta}\log\left[\left | \rho \right |\dfrac{\tau_0 \tau_0^*}{\tau_1 \tau_1^*}\right]\qquad$ (6)

After working for a while with (4), I could find this:

$Re(\Phi_1) - Im(\Phi_2) = \dfrac{1}{\beta}\tan^{-1}\left[i\dfrac{e^{-i\delta}(\tau_0 \tau_1^*)^{2} - e^{i\delta}(\tau_0^{*} \tau_1)^{2}}{e^{i\delta}(\tau_0 \tau_1^*)^{2} + e^{-i\delta}(\tau_0^{*} \tau_1)^{2}}\right]\quad$ (7)

But from here I don't know how to obtain $\Phi_1$ and $\Phi_2$, I did some other things, but this approach was the one that seemed most correct. I'm also not 100% sure that everything that I did is actually right. Am I missing something? Thank you.

LeviT
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  • Is $\tau_2$ a typo for $\tau_0$? – Gerry Myerson Sep 12 '23 at 09:44
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    Maybe start with $\tan\beta\Phi_1=(\sin\beta\Phi_1)/(\cos\beta\Phi_1)$, then $\sin z=(e^{iz}-e^{-iz})/(2i)$, $\cos z=(e^{iz}+e^{-iz})/2$, and so on. – Gerry Myerson Sep 12 '23 at 09:48
  • Yes, it was a typo. I fixed, thank you! – LeviT Sep 12 '23 at 13:27
  • $\Phi = \Phi_1 + i\Phi_2$, yet $\Phi_1,\Phi_2$ are complex numbers? There are infinitely many pairs of complex numbers $\Phi_1,\Phi_2$ that would produce the same $\Phi$. Surely you should have $\Phi_1,\Phi_2$ are real numbers. In the equations you are trying to prove, $\Phi_2$ is manifestly real. That $\Phi_1$ is real is not so clear, but just requiring $\Phi_2$ to be real is not enough to get a unique solution. – Paul Sinclair Sep 13 '23 at 11:58
  • The fact that $\Phi_1$ and $\Phi_2$ were also complex numbers really bothered me, but, after hours of trying, there is a chance that this question is poorly formulated. I tried solving with $\Phi_1$ and $\Phi_2$ being real numbers and got (6) for $\Phi_2$ and (7) for $\Phi_1$. I couldn't get the exactly equation of the statement, but I substituted the solution for $\Phi_2$ in (1) and got (7) for $\Phi_1$ again. Tomorrow I'll be able to contact the professor who brought this question and I think it will end all doubts. – LeviT Sep 14 '23 at 02:03

1 Answers1

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The question statement was wrong, $\Phi_1$ and $\Phi_2$ are real, so the conjugate of $e^{i\beta \Phi}$ is $e^{-i\beta \Phi_1}e^{-\beta \Phi_2}$. Also, the expression for $\Phi_1$ has the squares just like (7)

Using the same mathematical process, you can obtain $\Phi_1 = (7)$ and $\Phi_2 = (6)$. Thank you all for the help.

LeviT
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