# Does the characteristic polynomial contains all the poles of a closed-loop transfer function?

The Transfer Function of a Closed-Loop system is: $$T(s)=\frac{Z(s)}{1+G(s)Z(s)}$$.

The BIBO stability of $$T(s)$$ is guaranteed if all the poles of $$T(s)$$ have negative real value.

In many textbooks I've read, it's written without further explanation that the zeros of the characteristic polynomial $$1+G(s)Z(s)=0$$ are the poles of $$T(s)$$, without any conditions on $$Z(s)$$.

In one textbook, it was written that $$Z(s)$$ should be bounded. If $$Z(s)$$ is bounded, i.e. has no poles, it's obvious that the characteristic polynomial contains all the poles of a closed-loop transfer function.

But in many physical systems, $$Z(s)$$ has poles. I have a theoretical example in which $$Z(s)$$ has a pole at $$s_0$$, while $$G(s)$$ has a zero at $$s_0$$, such that their product is finite and different from 1 $$Z(s_0)G(s_0)\rightarrow A_0 \neq -1$$. Therefore, $$T(s)$$ will have a pole at $$s_0$$, while the characteristic polynomial will not have a root at $$s_0$$. Therefore, we found an example for characteristic polynomial that doesn't contains all the poles of a closed-loop transfer function.

Therefore, it seems as if a necessary condition for working only with the characteristic polynomial is that the $$Z(s)$$'s poles won't coincide with the $$G(s)$$'s zeros. But is there a necessary and sufficient condition?

Why is this question not addressed in control-theory textbooks? If it does, please refer me to those textbooks.