Yes, that works. Notice that you don't have to quote the fundamental theorem of algebra. If you found the zeros of the denominator you already got them. It doesn't matter if it is true or not that every polynomial has as many roots as its degree over the complex.
Even if you haven't found all of the roots you can still argue if a given (or already found) root is a simple pole. The idea is computing $B'(z_0)$ at that point. Remember that a zero $z_0$ of a polynomial $P$ is a zero of order $k$ iff $P(z_0)=...=P^{(k-1)}(z_0)=0$ and $P^{(k)}(z_0)\neq0$. So, in the process of using the formula you quoted you already are checking that the pole is simple.
$$\cot^2(z) = \frac{\cos^2(z)}{\sin^2(z)} \ \ \text{ so we need only worry about the zeros of } \sin^2(z)$$
These zeros occur when $z_k = k \pi$ for $k \in \mathbb{Z}$, as you've already identified.
Moreover, they are 2nd order zeros, as $(\sin^2(z))' = 2 \sin(z) \cos(z)$, which also has zeros at $z_k$, whereas $(2 \sin(z) \cos(z))' = 2 \cos^2(z) - 2\sin^2(z)$, which is nonzero at $z_k$.
Hence, $\cot^2(z)$ has $2^{nd}$ order poles when $z_k = k \pi, k \in \mathbb{Z}$. Therefore,
$$\text{Res}_{z = z_k} \cot^2(z) = \lim_{z \to z_k} \frac{d}{dz}\frac{\cos^2(z)(z - z_k)^2}{\sin^2(z)}$$
Calculate this derivative, then take the limit by applying L'Hopital's rule
In general, if $f(z)$ has an $n^{th}$ order pole at $z = z_k$, then
$$\text{Res}_{z = z_k} f(z) = \frac1{(n-1)!} \lim_{z \to z_k} \frac{d^{n-1}}{dz^{n-1}} \left[(z-z_k)^n\, f(z)\right]$$
Which can be seen directly from the Laurent expansion of $f$
Best Answer
The answer is that these are not poles - they are branch point singularities. They are not covered by the residue theorem, and if you were to include this function in a contour integral, you would want to point your contour so as to not include the branch cut $z \in [-r_0,0]$ inside the contour.