Poles and Zeros

From the quadratic formula, the two poles are located at

$$s = -\eta \pm \sqrt{\eta^2 - \omega_0^2} \;\isdef \; -\frac{1}{2RC} \pm \sqrt{\left(\frac{1}{2RC}\right)^2 - \frac{1}{LC}}$$

and there is a zero at $s=0$ and another at $s=\infty$. If the damping $R$ is sufficienly small so that $\eta^2 < \omega_0^2$, then the poles form a complex-conjugate pair:

$$s = -\eta \pm j\sqrt{\omega_0^2 - \eta^2}$$

Since $\eta = 1/(2RC) > 0$, the poles are always in the left-half plane, and hence the analog RLC filter is always stable. When the damping is zero, the poles go to the $j\omega$ axis:

$$s = \pm j\omega_0$$