How can i classify an analog filter given its pole zero map.

For example:
I've got my zero's located at $±2j$ and my poles located at $-1$ & $-2$ , then what is the nature of the filter?
$ i.e, $ It is L.P.F or H.P.F or B.P.F or B.S.F or A.P.F ?

My Approach:
Consider the following fig.:
enter image description here
Let P be a point on positive $j \Omega$ axis ,
$i.e,$ we are considering positive range of frequencies
Let pole at '$-1$' be denoted by $X_1$,
Let pole at '$-2$' be denoted by $X_2$,
Let zero at $2j$ be denoted as $Z_1$ and
Let zero at $-2j$ be denoted as $Z_2$

Now, at frequency $=0$ , P is at the origin (O) ,
Distance of the poles from P at this frequency is $PX_1 =1$ & $PX_2=2$
and distance of zero is $PZ_1=2$ $[\because \text{ we are considering +ve freq } , \therefore \text{ neglected } Z_2]$
So, Magnitude Response of the given filter at $\Omega=0$ is $$M= \frac{2}{1.2}=1 \quad \dots (i)$$ $\implies$ Pass Band at $\Omega=0$

Now, from the fig, we recognise as $\Omega$ increases from $0 \to 2$ , $M$ decreases ;
and at $\Omega=2$ , $M=0 \implies$ Stop Band at $\Omega=2$;
Now again as $\Omega$ increases from $2 \to \infty$ , $M$ increases ;
So, the nature of filter should be B.S.F or Notch filter whose notch frequency is at $\Omega_0=2 \quad \dots (ii)$

But, if we invoke its magnitude at $\Omega \to \infty$, then we get: $$M= \frac{PZ_1}{(PX_1).(PX_2)}=0 \text{ as } \Omega \to \infty$$ which contradicts $(ii) $ , so how to solve this?
I know it's the two zeroes that cancelling the effect of two poles at $\Omega \to \infty$ , but if we take into consideration of two zeroes then how will we approach $(i)$ ? Any help please...


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