$\cot(z)=\dfrac{\cos z}{\sin z}$ and I am supposed to find poles at $z=k\pi \quad k=0,\pm1,\pm2... $. But derivative of $\dfrac{d}{dz}\cot(z)=-{\csc}^2(x)$ and it is also singular at $z=k\pi$. So why do we identify poles of cot(z) as simple poles but not essential singularities?
Do you have any suggestions for reading on this topic ?
This is because $\sin k\pi=0$ giving $$z\cot z=\frac{z\cos z}{\sin z}=\frac{z\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots\right)}{z\left(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\cdots\right)}=\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}-\cdots\right)\left(1+\frac{z^2}{3!}-\frac{z^4}{5!}+\cdots\right)$$ so $z=k\pi$ is a pole of order one, or a simple pole.