What can be said about the poles of tan(1/z)?

Question

I came around the fact that $tan\frac{1}{z}$ has an essential singularity at $z=0$. I followed the following steps: $$tan(1/z)=\frac{sin(1/z)}{cos(1/z)} \\ cos(1/z) = 0 \implies 1/z=(2n+1)\pi/2 \implies z=2/\pi(2n+1) \\ $$ As $n$ tends to $\infty$, the sequence of zeroes of $cos(1/z)$ converge to $z=0$. Hence $z=0$ is not a isolated zero of $cos(1/z)$ and hence $z=0$ is an essential singularity of $tan(1/z)$.
It is also clear that $z=2/\pi(2n+1)$ are isolated singularities of $tan(1/z)$.
Now, the next thing I want to do is to prove that $z=2/\pi(2n+1)$ are poles of $tan(1/z)$ and they are simple. How do I proceed for it?

Answer 1

Actually, $0$ neither is nor isn't an essential singularity of $\tan\left(\frac1z\right)$. The classification of singularities into regular points, poles and essential singularities applyes only to isolated singularities, and $0$ is not an isolated singularity.

Now, if $z_0=\frac2{\pi(2n+1)}$ for some $n\in\Bbb N$, then $z_0$ is not a zero of $\sin\left(\frac1z\right)$ and it is a simple zero of $\cos\left(\frac1z\right)$ (since $\left(\cos\left(\frac1z\right)\right)'|_{z=z_0}\ne0$), and therefore $z_0$ is a simple pole of $\tan\left(\frac1z\right)$.