Given $\mu > 0$ and $X(y)\geq 0$ for all $y \in \mathbb{R}$, let us define \begin{equation} G(X(y)) = \frac{jX(y)+\mu}{jX(y)-\mu} \end{equation} I want to find an expression for $X(y)$ in terms of a given function $\Phi(y) \in [-\pi,\pi)$ such that the following holds: \begin{equation} \angle G(X(y)) = \Phi(y) \end{equation} Here is what I have done. Since we impose that $\mu > 0$ and $X(y) \geq 0$, we have [1, Eq.(1)]: \begin{align} \angle (jX(y)+\mu) &= \tan^{-1}\left(\frac{X(y)}{\mu}\right), \\ \angle (jX(y)-\mu) &= \tan^{-1}\left(-\frac{X(y)}{\mu}\right) + \pi \end{align} Therefore, \begin{align} \angle G(X(y)) &= \angle(jX(y)+\mu) - \angle(jX(y)-\mu)\\ &= \tan^{-1}\left(\frac{X(y)}{\mu}\right) - \left[\tan^{-1}\left(-\frac{X(y)}{\mu}\right) + \pi\right] \\ &= \tan^{-1}\left(\frac{\left(\frac{X(y)}{\mu}\right) - \left(-\frac{X(y)}{\mu}\right)}{1+\left(\frac{X(y)}{\mu}\right)\left(-\frac{X(y)}{\mu}\right)} \right) - \pi \\ &= \tan^{-1}\left( \frac{2\mu X(y)}{\mu^2 - [X(y)]^2} \right) - \pi \end{align} where the third line of the above equation follows from a well-known result on the difference of inverse tangents [2].
Due to the condition $\angle G(X(y)) = \Phi(y)$, we have \begin{align} \tan^{-1}\left( \frac{2\mu X(y)}{\mu^2 - [X(y)]^2} \right) = \Phi(y) + \pi \end{align}
Naturally, I would like to just take the tangent of both sides. This would give me a quadratic polynomial in terms of $X(y)$ and I can easily solve for $X(y)$, expressed in $\Phi(y)$. The problem is, the range of $\Phi(y)$ is $[-\pi,\pi]$ while $\tan^{-1}\left( \frac{2\mu X(y)}{\mu^2 - [X(y)]^2} \right)$ returns an angle that is in $[-\pi/2,\pi/2]$, so my idea does not work (as justified by the numerical results in MATLAB).
Does anyone have an idea on how to obtain a neat expression for $X(y)$ in terms of $\Phi(y)$? Especially, taking into account the implementation in MATLAB.
Thanks in advance.
[1] https://gubner.ece.wisc.edu/notes/MagnitudeAndPhaseOfComplexNumbers.pdf