I'm trying to analize this block, where a resistor and a capacitor share their terminals:

Obviously

\[ v_R = v_C = v \]

A specific condition may be found (in the Laplace domain) where the net current i = 0. This occurs when the current on the resistor is equal and with opposite sign with respect to the current on the capacitor. Maybe this is equivalent to stating that the capacitor is continuously charging and discharging to/from the resistor.

\[ i = i_R + i_C\\ i_R = -i_C \rightarrow i = 0 \]

This is verified for

\[ v_C = \frac{i_C}{sC}\\ i_C = sC v_C\\ i_C = -i_R\\ sCv_C = - \frac{v_C}{R}\\ s = - \frac{1}{RC} \]

which corresponds to the real frequency

\[ \omega = \left| - \frac{1}{RC} \right| \]

In this condition, the net current i through this block is 0, because i_C = -i_R, regardless of the rest of the circuit. This may represent a zero of this system, because if an external voltage or current is applied, the output current is 0.

In the Laplace domain it all seems ok, but I'm struggling to figure out how these currents may be equal and opposite in the time domain.

If for example (due to some external excitation)

\[ v_R (t) = V_0 \cos (\omega t + \phi) \rightarrow i_R(t) = \frac{V_0 \cos (\omega t + \phi)}{R} \]

we also have

\[ i_C (t) = C \frac{dv_C(t)}{dt} = C \frac{dv_R(t)}{dt} = -\omega V_0 C \sin (\omega t + \phi) \]

With

\[ \omega = \frac{1}{RC} \]

it is

\[ i_C (t) = C \frac{dv_C(t)}{dt} = C \frac{dv_R(t)}{dt} = - \frac{V_0}{R} \sin (\omega t + \phi) \]

This is not -i_R. They are simply two sinusoidal waves out of phase. They are not the opposite of each other. I can't figure out how, only at a specific frequency, it should be i_C = -i_R, and why.