Are all singularities of $\tan(z)$ simple poles?

Question

$\tan(z)$ has singularities at points $z=(2n+1)\frac{\pi}2$, where $n$ is an integer.

Are all these points of singularity a simple pole? Or do we need to obtain the type of singularity at each point by obtaining the Laurent's expansion about that point?

Answer 1

Here is another way to visualize all their poles are simple poles. Use the Mittag-Leffler expansion for meromorphic functions:

$$\begin{align}\tan(z)&=\sum_{n=0}^\infty \frac{8z}{(2n+1)^2\pi^2-4z^2}\\ \\ &=\sum_{n=0}^\infty \frac{2}{(2n+1)\pi-2z}-\frac{2}{(2n+1)\pi+2z}\\ \\ &=\sum_{n=-\infty}^\infty \frac{2}{(2n+1)\pi-2z}\\ \\ &=\sum_{n=-\infty}^\infty \frac{-1}{z-\left(n+\frac{1}2\right)\pi}\end{align}$$

All poles at $z_n=\left(n+\frac{1}2\right)\pi,~~n\in\mathbb{Z}$ are simple poles.

Answer 2

They are all simple poles because the poles are due to the cosine function (the sine is entire and is nonzero at all of the poles of the tangent), which only has simple zeroes. This last fact can be verified by, say, L'Hopital's rule. $$\lim_{z\to\frac{\pi}{2}(2n+1)}\frac{\cos(z)}{z-\frac{\pi}{2}(2n+1)}=(-1)^{n+1}\in\Bbb C\setminus\{0\}$$For $n\in\Bbb Z$.