I have two zeros at $z=-1$ and two complex conjugate poles at $z=A\cos\theta\pm jA\sin\theta$
This gives me the next transfer function
To get filter's DC gain I'm trying to substitute $z$ with $1$:
Following graphical method described here (gain as a product of lengths of vectors from every zero to given point divided by product of lengths of vectors from every pole to given point) I've got next formula:
Where is my mistake? Why I've lost square root in the first case?