Power Transfer Efficiency

I got lucky while solving this one. Let the current through the left $Z_C$ branch be $I_{C1}$ and that through the right $Z_C$ branch be $I_{C2}$ . The circuit equations by applying Kirchoff's current and voltage law are:

$Z_SI_S + I_{C1}Z_C = V_S$ $Z_LI_L + I_{C2}Z_C - I_{C1}Z_C=0$ $Z_RI_R + I_{C2}Z_C=V_R$ $I_S = I_L + I_{C1}$ $I_{C2} = I_L + I_R$

Rearranging into a matrix form gives:

$\left[ \begin{matrix} Z_S&0&0&Z_C&0\\0&Z_L&0&-Z_C&Z_C\\0&0&Z_R&0&Z_C\\1&-1&0&-1&0\\0&1&1&0&-1 \end{matrix}\right]\left[ \begin{matrix} I_S\\I_L\\I_R\\I_{C1}\\I_{C2} \end{matrix}\right]=\left[ \begin{matrix} V_S\\0\\V_R\\0\\0 \end{matrix}\right]$

$\implies \left[ \begin{matrix} I_S\\I_L\\I_R\\I_{C1}\\I_{C2} \end{matrix}\right] = \left[ \begin{matrix} Z_S&0&0&Z_C&0\\0&Z_L&0&-Z_C&Z_C\\0&0&Z_R&0&Z_C\\1&-1&0&-1&0\\0&1&1&0&-1 \end{matrix}\right]^{-1} \left[ \begin{matrix} V_S\\0\\V_R\\0\\0 \end{matrix}\right]$

Now, as per the link, the power supplied by the sources $S$ and $R$ are:

$P_S=I_S^* V_S$ $P_R=I_R^* V_R$

The superscript $*$ indicates complex conjugate, which is what I was not doing in my initial attempts.

It turns out that the real part of $P_S$ (active power component) is positive which means that power is supplied, and that of $P_R$ is negative which means that power is consumed. The required answer is, therefore:

$\boxed{\frac{-\mathrm{real}(P_R)}{\mathrm{real}(P_S)} \approx 0.9184}$