Why does $\tan(z)$ have a non-isolated singularity when z tends to infinity? My professor told me that there's a non-isolated singularity of $\tan(z)$ as $z$ tends to infinity.  However, I'm wondering why is this the case, since for $\tan(z)$ to have a singularity, $\cos(z)=0,$ which definitely are isolated singularities.
 A: Another way to look at this is to note that the behavior of a function $f: \bar{\mathbb C}  \to \bar{\mathbb C}$ as $z \to \infty$ is equivalent to considering $g(z) = f(1/z)$ as $z \to 0$.  For $f(z) = \tan z$, even just along the line $\Im(z) = 0$ we have singularities of $g$ whenever $z \in S = \left\{( \frac{\pi}{2} + k \pi)^{-1} : k \in \mathbb Z\right\}$.  But clearly, for any sufficiently small $\epsilon > 0$, there is always some $w$ satisfying $-\epsilon < w < \epsilon$ such that $w \in S$; i.e., no matter how small a radius we draw around the point $z = 0$, we can find a singularity of $g$ in that neighborhood.  This means the singularity at $0$ is not isolated, and the corresponding singularity of $f$ at $\infty$ is not isolated.
A: My guess is that what your teacher told you is that


*

*$\tan$ has a singularity at $\infty$;

*that singularity is not isolated.


The fact that it is not isolated is a consequence of the fact that $\tan$ has singularities at all the points of the form $\frac\pi2+k\pi$ ($k\in\mathbb Z$).
