Show that the function $f$ has pole of order 1 Problem:
Show that the holomorphic function $f$,
$$\frac{z-\pi}{(e^{iz}+1)^2}:D'(\pi;2\pi)\rightarrow \mathbb{C}$$
has a pole of order 1 at the point $\pi$.
This might be really simple but I cannot show that this has pole of order 1 but rather pole order 2, since the denominater is raised to the power of 2? Or am I wrong? When trying to show that it has a pole of order 1 I end with 0/0 but maybe this is sufficient and that 0/0 can be 1 if we compute the limit for $z\rightarrow\pi$?
I hope someone can clear this up for me. Notice that this is not homework. I'm just trying to get a better understanding of holomorphic functions. Thanks in advance.

The order of the numerator at $\pi$ is $1$, whereas the order of the denominator at $\pi$ is $2$. Therefore, the order of your function at $\pi$ is $-1(=1-2)$. In other words, your function has a simple pole at $\pi$.
