In the context of medical imaging, I need to solve the following equations for $\phi_1$ and $\phi_2$: \begin{alignat}{2} I_1(\boldsymbol{X}) &= \left\{A(\boldsymbol{X})+B(\boldsymbol{X})\cos(k_e X_1)\right\}e^{-i\phi_1(\boldsymbol{X})}e^{+i\phi_2(\boldsymbol{X})}\\ I_2(\boldsymbol{X}) &= \left\{A(\boldsymbol{X})-B(\boldsymbol{X})\cos(k_e X_1)\right\}e^{+i\phi_1(\boldsymbol{X})}e^{+i\phi_2(\boldsymbol{X})} \end{alignat} where $I_1$ and $I_2$ are MR images (known) and $\boldsymbol{X}=(X_1,X_2)$ denotes the tissue position.
I have tried playing with some algebra to obtain some estimations for both phases but cannot see how to proceed (see equations below).
\begin{alignat}{2} I_1+I_2 &= \left\{2A\cos(\phi_1)+2iB\cos(k_eX_1)\sin(\phi_1)\right\}e^{+i\phi_2(\boldsymbol{X})} \\ I_1-I_2 &= \left\{2iA\sin(\phi_1)+2B\cos(k_eX_1)\cos(\phi_1)\right\}e^{+i\phi_2(\boldsymbol{X})} \\ I_1+I_2^* &= \left\{2A\cos(\phi_2)+2iB\cos(k_eX_1)\sin(\phi_2)\right\}e^{+i\phi_1(\boldsymbol{X})} \\ I_1-I_2^* &= \left\{2iA\sin(\phi_2)+2B\cos(k_eX_1)\cos(\phi_2)\right\}e^{+i\phi_1(\boldsymbol{X})} \\ \end{alignat}
Anyone knows what would be the best way to solve it?
Edit 1: based on a comment, I have tried multiplying $I_1$ and $I_2$ (and their complex conjugates). The problem is that $\phi_1$ and $\phi_2$ are smooth in space, but the cosine modulation given by $\cos(k_e X_1)$ affects their smoothness. When $B(\boldsymbol{X})$ is small compared to $A(\boldsymbol{X})$, multiplying and taking the angle works nice but not for more general cases
Thanks in advance!
You have $$ e^{-i\phi_2} I_1=(A + B\cos(k_e X_1))e^{-i\phi_1}\\ e^{-i\phi_2} I_2=(A - B\cos(k_e X_1))e^{i\phi_1}\\ e^{-i\phi_2}(I_1 + I_2)= 2A\cos(\phi_1)\\ e^{-i\phi_2}=\cos(\phi_2)-i\sin(\phi_2)=\frac{2A\cos(\phi_1)}{(I_1 + I_2)} $$ but the r.h.s. is totally real which implies $\phi_2=n\pi $ and $\phi_1=\cos^{-1}(\frac{\cos(n\pi) (I_! + I_2))}{2A})$