I have to find the kind of singularity of the following functions:
$\tan^2(\pi z)$
cot$(z^2)$
What I have tried:
$\tan^2(\pi z) = \left( \frac{\sin(\pi z)}{\cos (\pi z)} \right)^2$. $\cos(\pi z) = 0 \iff z= \frac{1}{2}+2k\pi, k \in \mathbb{Z}$. Since we have a squared function, this is a pole of order 2.
cot$(z^2) = \frac{\cos (z^2)}{\sin (z^2)}$. $\sin(z^2) = 0 \iff z = \sqrt{k \pi}, k \in \mathbb{Z}$. So we have a pole ar $z = \sqrt{k \pi}$ of order 1
I don't know exactly why these are poles, and I don't know if I am missing some singularities. Can somebody explain me how to find the kind of singularities easily, or how to approach such problems?