Butterworth low pass filter zeros location after bilinear transformation explanation I am studying in a text book the transformation of a continuous time Butterworth low pass filter into a discrete time filter by means of bilinear transformation:
$$
    s = \frac{2}{T_d}*\frac{1-z^{-1}}{1+z^{-1}} 
$$
which imposes the following relationship between discrete and continuous frequency variables:
$$
     w = 2*\arctan(\frac{\Omega*T_d}{2}) 
$$



As can be clearly seen from this formulas a value of $ \Omega $ = inf causes w = pi and a value of $ \Omega $  = -inf causes w = -pi. However, both values of w correspond to z = -1 in the Z plane

In the example a 6th order continuous time Butterworth low pass filter is designed. Its magnitude squared function in continuous time is defined by:
$$
  |H_c(j\Omega)|^2 = \frac{1}{1 + (\frac{j\Omega}{j\Omega_c})^{2N}   }   
$$
From this formula it can be seen that as $\Omega$ goes to infinity the magnitude function goes to zero. So I can understand that in a 6th order filter there are 6 zeros in infinity (in S plane), which in discrete time means there are 6 zeros in w = pi and hence 6 zeros in z = -1. But:


What about the behavior of the system in s = -inf? 
Isn't the magnitude squared function symmetric ?


Wouldn't it also cause the magnitude function to go to zero and should generate zeros on w = -pi and z = -1
An $N^{th}$-order analog prototype system results after bilinear transform in an $N^{th}$-order discrete-time system. So the number of zeros and poles remains the same. All zeros at $|s|\rightarrow\infty$ map to $z=-1$. In your case there will be $6$ zeros at $z=-1$ because there are $6$ zeros at $|s|\rightarrow\infty$.

Of course, the magnitude (squared) response is symmetric, but the zeros at $|s|\rightarrow\infty$ cause the zeros in the magnitude at $\Omega\rightarrow\infty$ as well as at $\Omega\rightarrow -\infty$ (you can imagine those points meet at infinity). This becomes more obvious in the $z$-domain where moving towards positive frequencies means moving counterclockwise from $z=1$ (i.e., $\omega=0$), whereas moving towards negative frequencies means moving clockwise from $z=1$. In both cases you'll end up at $z=-1$.