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Although the input signal symbol oddly suggests a DC voltage, as well as both the inputs and outputs are at unusual places (in general: Input-left/Output-right), the first thing to do is to apply the Laplace to all the components:
R => R
XC => 1/sC
XL=> sL
And from then on apply kirchhoff's law to the nets in order to obtain circuit equations, then rearrange all parameters so that the Vs/Vin lies at the left side of equation. The denominator of that equation will give you the pole(s).
Anyway, if you want a simpler process, doing just by visual inspection, I would bet that there exist 1 pole at 2.pi.f.C2.R2 and 1 zero at 2.pi.f.C1.R3, perhaps subjected to some condition (e.g Rs >>R2), but you should confirm that with the above procedure.
A simple way to find poles in simple circuits, is to find the resistance each capacitor is "looking" into to. This is done by replace the given capacitor with a voltage or current source and then find the current or voltage for this source. The ratio is then the resistance the capacitor sees. The capacitor and found reistance multiplied is the timeconstant.
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