The given problem involves calculating the small signal input impedance \(Z_{in}(j\omega)\) in an NMOS transistor circuit with capacitors \(C_1\) and \(C_L\). The NMOS transistor is in saturation, and we need to find the equivalent impedance considering the trans-conductance \(g_m\).

Let's analyze the circuit step-by-step:

1. The impedance of a capacitor in the frequency domain is given by \(Z_C = \frac{1}{j\omega C}\). Therefore, for capacitors \(C_1\) and \(C_L\), their impedances are \(\frac{1}{j\omega C_1}\) and \(\frac{1}{j\omega C_L}\) respectively.
2. The controlled trans-conductance source due to \(g_m\) inverts the voltage, hence can be represented as a negative impedance which affects the input impedance at \(Z_{in}\).
3. The effective input impedance \(Z_{in}(j\omega)\) can be derived by combining these impedances:
   • The series combination of \(C_1\) and \(C_L\) results in an effective impedance of \(\frac{1}{j\omega C_1} + \frac{1}{j\omega C_L}\).
   • The negative impact of the trans-conductance is given by \(-\frac{g_m}{C_1 C_L \omega^2}\), which arises from the feedback through the transistor.
4. By summing these impedances, the expression for \(Z_{in}(j\omega)\) becomes:

\(Z_{in}(j\omega) = -\frac{g_m}{C_1 C_L \omega^2} + \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L}\)

This matches the given correct answer option: \(-\frac{g_m}{C_1C_L\omega^2} + \frac{1}{j\omega C_1} + \frac{1}{j\omega C_L}\).

Figure: Circuit used for analysis