= Solution
For <electrical resistance> $R_p>0$ in a <parallel circuit> with <inductance> $L>0$, add <electrical admittances> and invert:
$$
Z=\left(\frac1{R_p}+\frac1{i\omega L}\right)^{-1}
=\frac{R_p\omega^2L^2}{R_p^2+\omega^2L^2}
+i\omega\frac{R_p^2L}{R_p^2+\omega^2L^2}.
$$
Conjugate <impedance matching> to the <series RC impedance> requires
$$
\frac1{\omega C}=\frac{\omega R_p^2L}{R_p^2+\omega^2L^2},
\qquad R_p^2=\omega^2L(CR_p^2-L).
$$
It is possible at a finite positive <angular frequency> exactly when $CR_p^2>L$, in which case
$$
\boxed{\omega=\frac{R_p}{\sqrt{L(CR_p^2-L)}},\qquad R=\frac{L}{CR_p}.}
$$
The expression for $R$ follows by substituting this $\omega^2$ into the <real part> of $Z$. If $CR_p^2\leq L$, no such finite matching <angular frequency> exists. Under literal equality, both passive inductive configurations fail by the sign of their <imaginary parts>.
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